RussianCodeCup's blog

By RussianCodeCup, history, 10 months ago, In English,

First of all, congratulations to moejy0viiiiiv for winning, LHiC and jcccccccccccccccccccccsb for coming second and third. The Final Round proved to be quite challenging, and we are happy that the results are tight.

The round was prepared by Aksenov239, GShark, izban, manoprenko, niyaznigmatul, qwerty787788 and SpyCheese, supervisor andrewzta. Special thanks for this round goes to izban who is the author of problems D, E and F, and also gave a lot of important comments for all problems of the round.

Now let us move on to the editorial.

A. Set Theory

First let us prove that the answer is always YES.

Let us iterate over bj and check if summing it up with all ai values don't result in values that we already have. If no conflict is found, add the corresponding bj to B.

Let us give some estimate for the maximum element of B. The reason that we cannot include bj2, to B is the equality ai1 + bj1 = ai2 + bj2, so bj2 = bj1 - (ai2 - ai1). Each element of B forbids O(n2) values, so max(B) is O(n3). That means that the answer always exists for the given constraints.

Now let us speed up the test that we can add a number to B. Let us use an array bad, that marks the numbers that we are not able to include to B. When trying the value bj, we can add it to B if it is not marked in bad. Now the numbers that are equal to bj + ai1 - ai2 are forbidden, let us mark them in bad. The complexity is O(n3) for each test case.

B. Similar Words

Let us consider the following similarity graph: the vertices are the prefixes of the given words, two vertices are connected by an edge if the corresponding words are similar. Note that the set of vertices in this graph is the same as in the trie for the given words, so it doesn't exceed the sum of lengths of words.

Let us prove that the resulting graph is a forest. If two words are similar, let us call the shorter one the parent of the longer one. Each vertex now has at most one parent, and there are no cycles, so the graph is a set of trees — a forest.

Now the required set is the independent set in the constructed similarity graph, so we can use dynamic programming or greedy algorithm to find it.

There are two ways to construct the similarity graph.

Way 1. Hashes For each prefix find its hash, make a set of all hashes. Now for each prefix remove its first letter, check if such hash exists. If it does, connect them by an edge.

Way 2. Aho-Corasick

Let us build Aho-Corasick automaton for the given set of words. The vertices of the similarity graph are automaton vertices. An edge exists between two vertices if the suffix link from one of them goes to the other one, and their depths differ by exacly 1.

C. Eleventh Birthday

Let us use divisibility rule for eleven. The number is divisible by eleven if the sum of digits at odd positions is equal to the sum of digits at even positions modulo 11. So for each number on a card there are only two parameters that we care about: the sign interchanging sum of its digits with digits at odd positions positive and digits at even position negative, and the parity of its digit count.

Let us divide all cards to two groups: with even digit count and with odd digit count. Let us first put cards with numbers that have odd count of digits. Half of them (rounded up) will have their sign interchanging sum used as positive, other half as negative. Let us use dynamic programming to find the number of ways to sum them up to have a given sum modulo 11. The state includes the number of cards considered, the number of cards that are used as positive, and the current sum modulo 11. There are two transitions: take the current card as positive, and take it as negative.

If there are no cards with odd digit count, no matter how you order even digit count cards the result modulo 11 is the same. So the answer is either 0 or n!. In the other case each even digit count card can be used either as positive, or as negative, independent of the other cards. Use analogous dynamic programming to count the number of ways to get each possible sum modulo 11.

Finally, combine results for even and odd digit count cards, getting the total sum modulo 11 equal to 0.

Time complexity is O(n2).

D. Masha and Cactus

Let us use dynamic programming for a rooted tree and some data structures. Denote as fv the maximal total beauty of edges that have both ends in a subtree of v, such that if we add them all to the subtree it would be a cactus.

To calculate fv let us consider two cases: v belongs to some cycle, or it doesn't. If it doesn't belong to any cycle, fv is equal to the sum of fu for all children u of v.

If v belongs to a cycle, let us iterate over all possible cycles it can belong to. Such cycle is generated by an added edge (x, y) such that LCA(x, y) = v. Try all possible such edges and then temporarily delete a path from x to y from a tree, calculate the sum of fu for all u — roots of the isolated subtrees after the deletion of the path, and add it to the beauty of (x, y).

Now we have an O(nm) solution.

To speed up this solution let us use some data structures. First, we need to calculate LCA for all endpoints of the given edges, any fast enough standard algorithm is fine. The second thing to do is to be able to calculate the sum of fu for all subtrees after removing the path. To do it, use the following additional values: gu = fp - fu, where p is the parent of u, and sv = sum(fu), where u are the children of v.

Now the sum of fu for all subtrees after x - y path removal is the sum of the following values: sx, sy, sv - fx' - fy', the sum of gi for all i at [x, x'), the sum of gi for all i at [y, y'), where x' is the child of v that has x in its subtree, and y' is the child of v that has y in its subtree. We need some data structure for a tree that supports value change in a vertex and the sum for a path, range tree or Fenwick are fine. The complexity is O((n + m)log(n)).

E. Satellites

Any point X can be given by two angles α = XAB and β = XBA.

The point 2, β2) is in coverage area of a satellite at the point 1, β1), if α2 ≤ α1 and β2 ≤ β1.

Let two satellites that want to create a communication channel are at points 1, β1) and 2, β2). Repeater must be positioned in such point 0, β0), that α0 ≤ min1, α2) and β0 ≤ min1, β2). To make it harder for it to get to other satellites coverage area, it is reasonable to maximize α0 and β0: α0 = min1, α2), β0 = min1, β2).

Now let us move to a solution. Consider a query of type 3: you are given two satellites at points 1, β1) and 2, β2). You must check whether the point 0, β0) = (min1, α2), min1, β2)) is not in the coverage area of any other satellite. That means that among all satellites with α ≥ α0 the maximum value of β is smaller than β0.

The solution is offline, it considers all satellites from the test data and just turns them on and off. Sort all satellites by α. Each moment for each satellite we store its β if it exists, or  - ∞ if it doesn't. Let us build a range tree for maximum, and update values as satellites come and go. The query is a range maximum. To avoid considering the satellites from the query, change their values to  - ∞ before the query, and restore them afterwards.

The next thing to do is to get rid of floating point numbers. Instead of calculating the angle values, we will only compare them using cross product. Similarly, instead of storing β values in range tree, we will store indices and compare them by cross product in integers.

The final remark: we need to check that the point 0, β0) is not inside the planet. The point is inside the planet, if the angle AXB is obtuse, that can be checked by a scalar product of XA and XB. The point X can have non-integer coordinates, so we will not look for it, but will use colinear vectors from among those connecting satellite points to A and B.

F. To Play or not to Play

We give the main ideas of the solution, leaving proofs as an exercise.

Let us introduce coordinates y(x), where x = exp1 - exp2, and y = exp1 (exp1, exp2 — experience of the first and the second player, respectively). Let us proceed with time, and keep the set of possible states at this plane. It is a polygon.

Lemma 1: in the optimal solution if at some moment both players can play the game simultaneously, they should do so.

Now consider all moments of time, one after another. There are three transitions that modify the polygon:

  • The first player can play for t seconds. The new polygon is the Minkowski sum of the previous polygon and the degenerate polygon: segment with two vertices (0, 0) and (t, t).
  • The second player can play for t seconds. The new polygon is the Minkowski sum of the previous polygon and the segment with vertices (0, 0) and ( - t, 0).
  • Both players can play for t seconds. Now all points with x-coordinates [ - C;C] have 2t added to their y coordinate, and other points have t added to their y coordinate.

Let us now see how the polygon looks like. It is x-monotonous polygon (each line parallel to y-axis intersects it via a segment), the lower bound of this polygon is y = 0 if x ≤ 0 and y = x if x > 0. Let us see how the upper bound of the polygon looks like.

We want to prove that y-coordinate of the upper bound of the polygon doesn't decrease, and it only contains segments that move by vectors ( + 1, 0) and ( + 1,  + 1).

We attempt an induction, and the induction step is fine for the first two transitions. But the third transition can make the upper bound non-monotonous at a point x = C. To fix it, let us change the definition of our polygon.

Instead of storing all possible reachable points, we will keep larger set, that contains the original set, and for each of its point we can get maximal experience for the first player not greater than what we could originally get.

Lemma 2: if at some moment t we take two points P1 = (x1, y1) and P2 = (x2, y2) such that C ≤ x1 ≤ x2, and y1 ≥ y2, and our player start training from those states, the maximal experience for point P2 is not greater than for the point P1.

So we can expand our upper bound for x from [C; + inf) by a maximum value of the correct upper bound and y = y(C). The similar proof works for a line y = y(C) - (C - x) for x in ( - inf;C].

Now we have an upper bound as a polyline that contains ( + 1, 0) and ( + 1,  + 1) segments. All is left to do is to modify the polyline using the described transitions. This can be done using some appropriate data structure, treap for example. The solution works in O(nlog(n)).

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By RussianCodeCup, 10 months ago, translation, In English,

Hi, everyone!

On September 10, 2017 at 13:00 Moscow time the final round of Russian Code Cup 2017 will take place! The round will be 3 hours long. 55 participants of the final round will compete for prizes and glory, and we are eager to watch their competition at

And we have a nice surprise for the others: if you would like to apply your skills at our finals problems, come to after the Finals is over, at 16:35 Moscow time, the round featuring RCC Finals 2017 problems will take place, open for everybody.

Some notes:

  • Round will use ACM rules;
  • It will be unrated;
  • Problem difficulty will be close to Div 1 round;
  • We ask RCC finalists not to publish or discuss problems after the end of the Finals before the CF round ends, of course you shouldn't participate in CF round;
  • Judging machines at RCC and CF are different, "my solution passed/failed at CF, and it was different at RCC" is not a valid appeal.

UPD The official contest is over, congratulations to the winners! And good luck to CF Round participants.

UPD2 Editorial

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By RussianCodeCup, history, 14 months ago, translation, In English,

A. Small Numbers

First of all, find prime factorization of numbers a and b.

After that you need to notice that if ab is divisible by p2, (where p is a prime number), it is either possible to divide both a and b by p instantly, or you will need to perform one of the latter two operations first to move one of p factors to the other number, and then divide both by p.

Obviously, parity of occurrence of prime numbers in the multiple ab remains unchanged in all operations. Therefore we can either remove p factor completely from ab or leave it occurring only once. After removing all primes, lets say the ones left are p1, p2, ..., pn. Let us call multiple of all these numbers d. Important observation is that n can't exceed 14 because multiple of first 15 prime numbers is more than 1018, which means it can't be a product of two numbers a, b ≤ 109. Now we simply need to iterate over all possible pairs and chose one with minimum sum that we can obtain from d. This can be done in O(2n) which fits the time limit.

B. New Keyboard

Let us use dynamic programming. The state is d[i][j][k], where i — is the flag that denotes the previous action (0 for layout switch, and 1 for typing character), j is the number of the current layout, and k is the number of characters typed so far. The value is the minimal time to reach the state.

Now iterate over k. For a given k first iterate j from 1 to n twice. Both times relax d[0][j % n + 1][k] = min(d[0][j % n + 1][k], min(d[0][j][k] + b, d[1][j][k] + a)). Two iterations of j from 1 to n are needed to ensure that if the layout switches from n to 1 it is processed correctly.

Now iterate j from 1 to n again and relax values for k + 1. If there is the k-th character of the message in the j-th layout, update d[1][j][k + 1] = min(d[0][j][k], d[1][j][k]) + c.

The answer is min(d[1][j][m]), where m = length(s), for all j from 1 to n.

C. Folding the Figure

Note that there are exactly 4 possible folding lines: two horizontal and two vertical, because the figure must be at one side of the folding line, and must touch it.

Take any square of the folded figure with the minimal x-coordinate. Let it be the cell (xi, yi). Take line x = xi as the folding line. Now we need to find k - n squares of the original figure on the other side of the folding line. So let us find k - n connected squares of the folded figure containing the square (xi, yi), for example, using DFS.

D. Acute Triangles

To count the number of acute triangles let us count the total number of triangles and subtract the number of right and obtuse triangles. For the purpose of this problem let us consider three points on a line to be a degenerate obtuse triangle.

The total number of triangles is equal to the number of ways to choose 3 points of n.

Now note the fact: each right or obtuse triangle has exactly one right or obtuse angle. So the number of bad triangles is equal to the number of bad angles.

Now let us count the number of angles not less than 90 degrees with vertices in the given points. Consider the angle vertex and sort other points by their polar angle relative to the chosen point. Now use two pointers: consider the first of the other two points, the matching third points form a continuous segment along the circle, and its ends move in the same direction.

Time complexity is O(n2log(n)).

E. Joining Arrays

Let us consider two solutions for the problem, that take O(k2·log(k)) and O(k2) respectively. The first one brings up core ideas, while the second one being more elusive has simpler implementation.

O(k2·log(k)) solution:

Consider three main steps of the solution

  1. For each array X (A or B) and each length 1 ≤ length ≤ |X| find minSubsequenceX[length] — lexicographically smallest subsequence of X that has the given length;
  2. Iterate over t such that 1 ≤ t ≤ min(k - 1, |A|) and 1 ≤ k - t ≤ |B|, take minSubsequenceA[t] and minSubsequenceB[k - t], join them;
  3. Joining the given sequences, get the optimal sequence of length k, update the answer with that sequence.

1) To find minSubsequenceX[length] for each length, let us do the following:

  • Calculate next[i][c] that will store the next occurence of value c after i in X;
  • Calculate firstSymbol[length][i] — the first character of lexicographically smallest subsequence of X[i..|X| - 1] that has given length. To calculate it note the following:
    • If j1 = next[i][1], and it exists, then firstSymbol[1][i], firstSymbol[2][i], ... firstSymbol[|X| - j1][i] are equal to 1;
    • If, if j2 = next[i][2], and it exists, then firstSymbol[|X| - j1 + 1][i], ..., firstSymbol[|X| - j2][i] are equal to 2;
    • ...
    • If j3000 = next[i][3000], and it exists, then firstSymbol[max(|X| - j1, |X| - j2, ..., |X| - j3000 - 1) + 1][i], ..., firstSymbol[|X| - j|alphabet|][i] are equal to 3000.
  • After that we can use firstSymbol[length][i] to restore lexicographically smallest subsequence of each array, one by one.

This step takes O(|X|2).

3) Given two lexicographically minimal subsequences SA and SB, now we need to join them to lexicographically smallest sequence. Let us use two pointers p1 and p2. If SAp1 ≠ SBp2, we move the smaller pointer, appending the value it points at to the answer. If SAp1 = SBp2, use binary search to find the longest common prefix of SA[p1..|SA|] and SB[p2..|SB|], and compare the following elements. Use hashes to compare subarrays of SA and SB.

This step takes O((|SA| + |SB|)·log(max(|SA|, |SB|))) = O(k·log(k)).

So the total time complexity is O(|A|2 + |B|2 + k2·log(k)) = O(k2·log(k)).

O(k2) solution:

Let us index arrays, let array A have number 0, and array B have number 1. Let us build the answer one element after another, and maintain the values dp[i][j], where i is the number of the array (0 or 1), j is the index in this array, dp[i][j] is the minimal index in array 1 - i, that can be appended to the answer, if we are at index j in array i.

At the t-th of the k iterations we find the minimum element that can be appended to the answer, and still the answer can be completed by adding another k - t - 1 elements. Also we must note that both subsequences from the two arrays must be non-empty.

After adding the element v, update the dp values in O(|A| + |B|). To do it, use next array, the same as in previous solution.

F. Two Trees

There is a requirement that k-subtree must have a vertex at depth k. Let us temporarily remove this limitation.

Consider all k-subtrees for some value of k. They can be divided to equivalence classes. Let each vertex v have ck[v] — the label of its equivalence class that its k-subtree belongs to.

If k = 0 all c0[v] are the same, because 0-subtree of any vertex is this vertex by itself.

If k = 1 then c1[v] is the number of children of v .

Now let us see how we can convert arrays ck[v] and cm[v] to an array ck + m[v]. First let us assign the array arrk + m[v] to each vertex v, that will uniquely identify its k-subtree. Let u1, ..., us be its descendants of level k in DFS order. Then arrk + m[v] = ck[v], cm[u1], ..., cm[us]. So its (k + m)-subtree is identified by its k-subtree and m-subtrees from the bottom vertices of its k-subtree. See the picture below for k = 3 and m = 1.

To get the list of descendants at level k, let us run one DFS from the root, when entering the vertex, put it to its k levels ancestor list.

To transform arrk + m[v] to integers ck + m[v] we can either use hashing, or trie, or unordered_map. Each vertex is considered only once in each of these arrays, so the total time is O(n).

Given array ck[v] it is easy to check whether there are two equal k-subtrees. To do it you must find two vertices with the same ck, considering only vertices that have descendants of level k (now we need to bring back this requirement).

To find the maximal such k, let us first count c1[v], c2[v], ..., c2t[v] (2t is the maximal power of 2 not exceeding n). After that make some kind of binary ascending by k: start with k = 0, and try to add 2t, 2t - 1, ..., 20, one by one.

Time complexity: O(nlog(n)).

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By RussianCodeCup, history, 14 months ago, translation, In English,

Ready, steady, go!

It's time to find out who the top 50 participants to compete in the Final Round of Russian Code Cup 2017 are. We invite everyone qualified to take part in the Elimination Round this Sunday, May 14, 2017 at 13-00 Moscow Time. Round length is 2 hours. We invite you to, and wish you good luck!

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By RussianCodeCup, history, 15 months ago, translation, In English,

A. Spreadsheets

Let us subtract 1 from k, now the columns are numbered from 0.

Let us first find out how many characters are there in the name of the column. To do it let us subtract powers of 26 from k, one by one, until the current power is greater than the remaining number k'.

Now the name of the column is k' in 26-based notation where characters A–Z are used as digits, prepended with leading A-s to the required length. You can either convert it using standard library function (in this case you must replace digits 0–9, A–P that will be used by the required digits), or implement a textbook algorithm of conversion to another base.

Final note: a day before the round one of the testers pointed out that CF Beta 1 Round problem B was similar to this one. After a discussion, taking into account that CF1 round was long ago, the problem is actually easier than CF1B problem, and we had no well prepared and easy enough replacement problem, we decided to keep the problem.

B. Mortal Combat

Let us count how many hit points the boss loses if attacked by the i-th hero: pi = ai·⌈ hi / A⌉. Sort heroes by non-increasing of pi and send them to attack the boss in this order.

C. A Bit Palindromic Numbers

Let us find out how to count a bit palindromic numbers among the first n positive integers. Now to solve the problem for the range from l to r we count the number of a bit palindromic numbers among first r, among the first l - 1, and subtract the latter from the former.

All numbers from 1 to 9 are a bit palindromic, so if n ≤ 9 the answer is n. Otherwise let us add 9 to the answer and count a bit palindromic numbers from 10 to n. Let us consider ten integers from 10k to 10k + 9, where k ≠ 0. Only one of them is a bit palindromic, the one that has the same first digit as k. So there is one a bit palindromic number from 10 to 19, one from 20 to 29, and so on.

Let n = 10k + d, where 0 ≤ d ≤ 9. So if d is greater or equal to the first digit of k, there are k a bit palindromic numbers from 10 to n, if it is less then the first digit of k, there are k - 1 such numbers.

D. Tree and Polynomials

Let us walk through the key ideas of the solution.

First, notice that6 you can solve problem independently for the two types of the queries. That is because operations don't depend on the values in the vertices.

Second idea is the following. Instead of storing particular values in the vertices, let us store polynomial that must be evaluated from it depth to get the value.

We cannot process queries directly, considering all vertices affected by a query, and adding the query polynomial to the polynomial in the vertex: that would take O(n2·k) which is too slow. Let us use lazy propagation then.

For the queries of the first type we need to add a polynomial q(t) to each vertex in a subtree of v. Let us just store information about that: for each vertex v let us store push1[v] that has the sum of all polynomials that affected the vertex v in queries of the first type.

Simultaneously let us store the sum of all affecting queries of the second type as a polynomial push2[v] for each vertex v.

Now denote as sum1[u] the sum of all polynomials that affect vertex u in queries of the first type.

To calculate sum1[u] let us run DFS from the root, and keep track of a polynomial cur. Entering the vertex v add push1[v] to cur, leaving v subtract it back. When we have entered the vertex u, assign sum1[u] = cur.

Similarly, denote as sum2[u] the sum of all polynomials that affect the vertex u in queries of the second type.

To calculate sum2[u] also use DFS, sum2[u] is equal to the sum of values sum2[v] for all children of u, and the value push2[u].

After we have calculated polynomials sum1[u] and sum2[u] for each vertex, evaluate them on d[u] and add the results, that is the answer for that vertex.

Time and memory are both O(nk).

E. Nice Report

Let's solve one additional problem before proceeding to the main one. You are given a sequence (a1, a2, ..., an) of some objects and somebody asks you to find out how many different objects are presented in this sequence. However, some restrictions apply to the ways you are allowed to process given data:

  1. once the object ai is read, it's lost forever: the next read will return ai + 1 and there is no way to obtain ai again;
  2. you can only use O(1) additional memory

To solve this subproblem we will use the fact that expected value of minimum of m value uniformly distributed over segment [0;1] is 1 / (m + 1).

Consider a hash function h: A → [0;1] which transforms objects ai to some number from [0;1]. Define M as M = min {h(a1), h(a2), ..., h(an)}. This value can be transformed to the problem solution as 1 / M - 1. However, we could run out of luck and end up in the situation where some h(ai) is too small and spoils the answer too much. To deal with it, we can average the answer over several runs with different hash functions h1, h2, ..., hk for some fixed k.

First, let's solve the case of acyclic graph. Any acyclic graph has topological sort: vertices order that respects edge directions: only edges from latter vertices to earlier ones exist. Let's label each vertex with some random number uniformly distributed over [0;1]. Topological sort existence allows us to calculate the minimal reachable label Mi for each vertex by some kind of dynamic programming. After this calculation, we can approximate the number of reachable vertices from i as 1 / Mi - 1. Again, we need to average results from several iterations to improve answer precision.

But given graph can contain cycles. To create an acyclic graph we can build the graph condensation by compressing strongly connectivity components. In the condensed graph each vertex i corresponds to wi vertices in the original graph. To finally solve the problem we'll change vertex labeling phase a bit: a simple random label becames minimal value from wi random value from [0;1] rolls.

A curious reader can find more information related to this problem by reading an article "Size-Estimation Framework with Applications to Transitive Closure and Reachability" located at

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By RussianCodeCup, history, 15 months ago, translation, In English,

Hi, everyone!

There are already more than 400 participants who are preparing for the Elimination Round of Russian Code Cup 2017. If you are still among those who has not qualified yet, we invite you to take part in the Third Qualification Round that will take part this Saturday, April 29, at 14-00 Moscow Time. Top 200 participants will also qualify to the Elimination Round and will be able to compete to get to the Final Round of Russian Code Cup 2017.

Good luck and see you at!

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By RussianCodeCup, history, 15 months ago, translation, In English,

A. Very Important Persons

There are two main ways to solve the problem.

The first way is to assign guests to seats by diagonals, starting from the seat (1, 1). You should be careful with considering cases n < m and m < n when implementing traversal.

The second way is a more common solution which however requires more advanced algorithm. Let us run BFS from (1, 1) and assign seats to guests in order they are popped from the queue, starting from the most important one.

B. Least Common Multiple

Let p / q be divisible by both a / b and c / d, all fractions are irreducible. So the numbers (p / q): (a / b) = (p·b) / (q·a) and (p / q): (c / d) = (p·d) / (q·c) are integers.

Since p·b is divisible by q·a, b and a are relatively prime, then p is divisible by a. p and q are also relatively prime, so b is divisible by q. Similar argument allows to conclude that p is divisible by с, and d is divisible by q.

Therefore p is divisible by lcm(a, c), q is a divisor of gcd(b, d). The fraction lcm(a, c) / gcd(b, d) is divisible by both a / b and c / d, and is therefore the smallest such fraction. So the answer is lcm(a, c) / gcd(b, d).

C. Bad Order

You have to add numbers to the given array to convert it to a permutation that requires maximal number of swaps when being sorted by selection sort.

First suppose that the permutation is fixed. Let us find the number of swaps that selection sort makes. Consider permutation as a set of cycles. Let us consider a cycle that contains the minimal number that occupies wrong position. Length of this cycle is greater than 1. After we make a swap, it is split to two cycles: of length 1, and 1 shorter than its initial length. Therefore the cycle of length L will be split L - 1 times, so the number of swaps required is n - c, where c is the number of cycles.

Therefore we must add numbers to the given array in such way that the number of cycles in the resulting permutation was minimal possible. To do it consider the given array as a directed graph, if a[i] = j, add an edge from i to j. The graph is a union of cycles and paths. All cycles in the graph will remain there after we convert the array to a permutation. All paths can be concatenated to a single long cycle (vertices without both incoming and outgoing edges are considered paths of length 0). To do so, order paths arbitrarily, and add an edge from the end of each path to the beginning of the next path. Finally make them a single cycle by adding an edge from the end of the last path to the beginning of the first one.

D. Red-Black Tree

First you must notice a subtle hint in the first sentence of the statement. It is easy to prove (or find in any data structures textbook) that red-black tree with n vertices has O(log(n)) black vertices on any path from the root to a leaf. Let us use this fact in our solution.

Let us call a tree almost red-black if its coloring satisfies all constraints of the red-black tree, except that its root is red.

We can now use dynamic programming to calculate for each subtree the number of ways to color it to become red-black or almost red-black, and to have the number of black vertices on any path from its root to a leaf equal to h.

Let us denote the described number of ways as d[v][h][t], here v is the vertex, h is the number of black vertices on any path from v to a leaf in its subtree, t is the type of coloring, t equal to 0 would, for example, mean that the subtree is red-black, and t equal 1 would mean that the subtree is almost red-black.

Now d[v][h][0] is the product for all u children of v of values (d[u][h - 1][0] + d[u][h - 1][1]). Similarly, d[v][h][1] is the product for all u children of v of values d[u][h][0].

The answer to the problem is the sum for all h of values d[1][h][0].

Since h for all valid colorings is limited by C = O(log(n)), we can consider the number of colorings be zero if h > C.

So we only need O(nlog(n)) states, and O(1) to calculate the result for each state. The time complexity is O(nlog(n)).

E. Array Study

Since there are the same constrains to n and q, let us use n instead of max(n, q). Let us make some observations.

Observation one: the answer to the query is the maximal length of [L, R], such that prefL - 1 = prefR, where prefi = a1 + a2 + ... + ai is the prefix sum of the given array. Let us move from the initial array to the array of prefix sums and for a query [l, r] look for the longest subarray of [l - 1, r] such that prefL = prefR.

Observation two: prefi are quite small ( - n ≤ prefi ≤ n).

Let us make use of queries given offline. Our method will be similar to Mo's algorithm. Split all queries to groups, the i-th group will contain queries [li, ri] such that i·K ≤ li < (i + 1)·K (here K is approximately sqrt(n)). In each group sort queries by their right end. Let us solve the problem separately for each group.

Consider the i-th group, where li is in [Li, Ri]. Let us consider queries by non-decreasing of ri, and maintain two arrays: mostLeft[p] and mostRight[p] — the first after Ri and the last before r occurrence of the prefix sum p. Using these two values, we can find the answer that is subarray of [Ri + 1, r]. To get the complete answer to [l, r] query, let us calculate the answer with the beginning of the required subarray at [l, min(r, Ri)] using mostRight array, and take maximum of that value and the optimal subarray for [R + 1, r].

So it takes O(K·ci + n) to answer all queries of one group, here ci is the number of queries in that group. Adding up all O(n / K) groups, we get the total time complexity O(sum(i = 1... n / K: K·ci + n)) = O(K·sum(ci) + n2 / K). Setting K = sqrt(n) we get O(n·sqrt(n)) complexity.

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By RussianCodeCup, 15 months ago, translation, In English,

Hello everyone!

Top 200 from the First Qualification Round have already advanced to the Elimination Round, but the other paricipants still strive to get there. If you haven't qualified yet, we invite you to take part in the Second Qualification Round that will take place this Sunday, April 16, at 12-00 Moscow Time. Best 200 participants will advance to the Elimination Round, other participants can try again in the Third Qualification Round.

Good luck to everyone, see you at!

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By RussianCodeCup, history, 16 months ago, translation, In English,

A. Martian Volleyball

First note that in order to end the game as soon as possible only the currently leading team must get points.

Consider the condition that one of the teams wins. It must score at least k points and have at least 2 points advantage.

So the answer is max(k, min(x, y) + 2) - max(x, y).

B. Painting the Wall

First note that if the maximal length of the continuous segment is L, at least L colors are needed to paint the wall. On the other side L colors is always enough.

Consider a square L × L, fill the first line with integers from 1 to L, fill the following lines with all its possible circular shifts. Use these squares to cover the given wall rectangle like this (L = 4):

1 2 3 4 1 2 3 4 1 2

2 3 4 1 2 3 4 1 2 3

3 4 1 2 3 4 1 2 3 4


Now each vertical and horizontal segment of length L or less has all tiles of different colors. All that is left is to bring back the lamps.

C. Magic Artifact

Let us denote advantage that Maxim gets if he finds artifact as ci, so ci = ai - bi.

Let Maxim first complete levels in order 1, 2, ..., n. If he finds the artifact at the i-th level, the time he needs to complete the game is equal to bj plus c1 + ... + ci. So the expected time of completing the game is:

b1 + ... + bn + p1·c1 + p2·(c1 + c2) + ... + pn·(c1 + ... + cn).

b1 + ... + bn doesn't depend on level order, so let us try to make the rest as small as possible. Expanding the sums, we get sum of all ci·pk for such i, k that i ≤ k.

Let us now swap levels i and i + 1 in our order. For the given i among ci·pk terms only the one equal to ci·pi + 1 changes — it is replaced with ci + 1·pi. If the order was optimal, it must be ci·pi + 1 ≤ ci + 1·pi, in the other case it was optimal to swap them and get a better answer. Transform this to:

ci / pi ≤ ci + 1 / pi + 1.

In the optimal answer for all i from 1 to n - 1 this inequality must be satisfied. Therefore it is optimal to sort levels according to the key ci / pi. Note that if pi = 0, you may consider ci / pi = ∞.

Sorting according to the key ci / pi should be done carefully. If the fractions are compared using floating-point division, the case ci = pi = 0 will lead to error because 0 / 0 = NaN. If the fractions are compared using the comparator ci·pk < ck·pi, the case ci = pi = 0 will also lead to error because the result of that comparison will always be false. That means that levels with ci = pi = 0 should be handled separately. Such levels can be completed at any time because they do not affect the result besides fixed terms bi. For example, they can be placed at the end.

After sorting the desired expected value can be found using the formula above, use prefix sums for ci to get it fast.

D. Memory Manager

First for each query qi let us find goi — the minimal j, such that among qj, qj + 1, ..., qi there are at most k different blocks — that would mean that it is possible to place pointers before the j-th query in such way that queries j, j + 1, ..., i are performed immediately. This can be made in O(sum(|qi|)), using, for example, two pointers.

Now let us use dynamic programming, denote as dpi the minimal time needed to perform the first i queries. If goi = 0 then dpi = 0. If the equation is not satisfied, dpi = minj = goi..i(dpj - 1 + sj) — we choose the query j to perform the pointer movement before. Since values goi do not descend, this value can be maintained using either std::set of dpj - 1 + sj, or queue with minimum query support.


First let us consider the problem for given two strings: to solve it calculate the number of occurrences of each letter in the first string, the number of occurrences of each letter in the second string, not including the first letter of the first string and the last letter of the second string, that must be used any way. Now the answer is the product of string lengths, decreased by the sum of values cnt1[letter] * cnt2[letter] for each letter. Let us leave proof as an exercise, such problem was at Northern Subregional Contest of NEERC 2015, Problem C. Unlimited participants could solve this problem at Three Quaterfinals Cup.

For n strings similar ideas can be used. Put all strings to a prefix trie, all strings to a suffix trie, the total number of ways to get a string is the product of the number of vertices in these tries. What we need to subtract is similar product of number of occurrences of letters in tries, first/last letters of each word must not be considered.

Now we need to answer segment queries. Let us make sqrt-decomposition, divide all strings to groups based on sum of their lengths. Let us greedily add strings to a group until their sum of lengths is more than sqrt(total sum of lengths). Note that the last string can have a big length.

Now use Mo's algorithm, sort queries using the block of the left end as a key, move right end only forward between left end block changes. To add/remove a string to a trie one must maintain the number of active vertices in a trie subtree for each vertex, and the number of entries of letters to a trie.

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By RussianCodeCup, 16 months ago, translation, In English,

Hi, everyone!

This Sunday, April, 2 at 19-00 Moscow time the First Qualification Round of Russian Code Cup 2017 will take place. 200 top participants will qualify for the Elimination Round, those who wouldn't qualify can still take part in two following qualification rounds.

We have a news for you: we have added Kotlin, Haskell and Free Pascal to our list of programming languages, also you can now submit your Python programs to run in PyPy. Exact versions of compilers and compilation commands are published at the official web site. We are working on adding more programming languages.

Good luck to everyone and see you at!

UPD Editorial is published

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By RussianCodeCup, 16 months ago, translation, In English,

Hello, everyone and thank you for participation!

First of all we would like to say sorry for the length of the testing queue during the first hour. We have selected IOIP problems for the Warmup Round — there are many common problemsetters for IOIP and RCC — the round was prepared by Victoria Erokhina (viktoria), Dmitry Filippov (DimaPhil), Stanislav Naumov (josdas), Mikhail Putilin (SpyCheese), Grigory Shovkoplyas (GShark), Andrew Stankevich (andrewzta), Ilya Zban (izban). We wanted to let everyone solve our interesting problems.

But one small glitch lead to a big failure. You might have noticed that we usually make multiple tests in one input data for RCC. That is because the most expensive operation is to run user solution. But IOIP has fewer participants, so the easiest problem had 32 tests. It is not too many for a problem, but during 15 minutes 400 participants submitted the correct solution and our testing machines capacity was not enough.

We decreased the number of tests in the problem, and the situation got better. We will take this into account when preparing problems for future contests.

Now let us proceed to problem analysis.

A. Spaceship

Note that if the last enemy to destroy has power equal to x, the sum of powers of all enemies is 2x. Therefore to find the power of the enemy to destroy last, let us find the sum of all powers and divide it by two. After that just pick any enemy with such power and swap it with the last one.

B. Rangers in the Bus

Let us process passengers one by one. Since Pink Ranger could take any seat, any passenger can be Pink Ranger.

For each seat we want to be able to quickly answer if it is free. Let us use std::set that stores the set of occupied seats. To check whether the current passenger can be Red Ranger let us find the seat that Red Ranger would choose. Iterate row from 1 to n, and if there is a free seat in the current row, take the left seat if it is free, or the right seat in the other case. If the current passenger chose that seat, he could be Red Ranger. Similarly we can check whether the current passenger could be Blue, Black or Yellow Ranger. After processing the passenger, add the seat he chose to the set of occupied seats.

The solution described uses O(nk) time and O(k) memory.

To get rid of time limit, let us note that all we need is to store the minimal and the maximal row that have free seats. Use two indices first and last to store them. Initially first = 1, last = n. After each passenger update these values. Increase first by 1, until the row pointed by first is completely occupied, then similarly decreaser last. There are at most k / 2 occupied rows, so both indices will be changed O(k) times.

C. Magic Weapon

Precalculate arrays: R[a][b] — the number of red details that have the first digit of the model number equal to a and the last digit equal to b; G[a] — the number of green details that have the last digit equal to a; и B[b] — the number of blue details that have the first digit equal to b.

It model numbers of different colors were distinct, the answer would be equal to the sum for all a and b of values G[aR[a][bB[b].

To resolve the situation when some pair of details have the same model number, calculate the number of red-blue pairs with the same model number, red-green pairs and green-blue pairs. Note that only model numbers with the same first and last digit need to be considered. You can use std::map to calculate the number of such model numbers. Now use inclusion-exclusion principle to find the answer: subtract the number of pairs of each type and add back twice the number of triples that all three rangers use detail with the same model number.

D. Knights and Knaves

Let us use dynamic programming with mask of the last two columns as a state. Consider for example subtask of calculating the maximal number of knights.

Let dp[i][mask_prev][mask_cur] be the maximal number of knights that can be positined among the first i columns, where mask_cur is the mask of the i-th column, mask_prev is the mask of the (i - 1)-th. Use bit equal to 1 to denote a knight and 0 to denote a knave.

Try all possible masks for the (i + 1)-th column, check whether constraints for the soldiers in the i-th column are satisfied, because now we know all of their neighbors.

The first and the last columns must be considered separately, because soldiers there don't have one of the neighbors.

Initial values: dp[2][mask_prev][mask_cur] the number of ones in mask_prev and mask_cur, if mask_prev can be before mask_cur.

Updating values: relax dp[i + 1][mask_cur][mask_next] with dp[i][mask_prev][mask_cur] + ones(mask_next), where ones(x) is the number of ones in x.

The answer is the maximum among dp[k][mask_prev][mask_cur], such that mask_cur can be after mask_prev.

Finally, we probably need to consider k = 1 separately, because there is no previsous column for any column in this case.

E. Parallelepiped

Let us give a sketch of the main solution idea. First, separately consider all parallelepipeds that have two or three equal sides. This can be done in O(n).

Now let us consider the case where all three sides are different. Let us build the following undirected graph: vertices are side length, connect a and b if there are at least two sheets of size a × b. Now the problem is reduced to considering all triangles in this graph which can be done in O(n2 / w) or O(nsqrt(n)) time (here w is the word size, 32 or 64 which comes up from bit compression).

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By RussianCodeCup, 16 months ago, translation, In English,

Hi, everyone!

We are pleased to announce Russian Code Cup 2017! Qualification rounds will start in April, the Final round will be held online in September. Each of the three Qualification rounds will allow 200 participants to qualify for Elimination round, top 50 from Elimination round will proceed to Final round. The problems will be both in Russian and in English.

The full schedule of the tournament can be found at, you can also register there. Note that you must register for the new season even if you took part in the previous tournaments.

We still have more than two weeks before the first Qualification, so we would like to invite everyone to the warm up round that will take plance on Sunday, March 19, 14-00 Moscow time. Visit to take part in the round, it will be 2 hours long.

Good luck to everyone and see you at Russian Code Cup 2017!

UPD: We remind you that the Warmup Round will start at 14-00 Moscow time. Problems of the Warmup Round are based on IOIP problems (Russian contest for high school students, run today at several cities) that are prepared by the same team of authors. We kindly ask students that take part in IOIP to play fairly and not to try to take part in RCC Warmup. Good luck to everyone!

UPD2: Thanks to all participants that made a real stress testing for our judging machines. Unfortunately we had long queues during the first hour, they were fixed, and from the middle of the contest the waiting time in most cases was limited to 2-3 minutes. We will make sure there would be no queues during further rounds.

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By RussianCodeCup, history, 22 months ago, translation, In English,

Hi, everyone!

Russian Code Cup 2016 is over, the problems for the Final Round were more difficult than usually, but the participants didn't give up. The champion is Gennady Korotkevich tourist, the second place winner is Vladislav Epifanov vepifanov, the third place winner is Nikolay Kalinin KAN. Congratulations to the winners, the complete results are at the RCC web site

Before proceeding to the tutorial, we would like to thank Mail.Ru Group for organizing the tournament and providing the prizes, and the judges from ITMO University for problems. Chief judge Andrew Stankevich andrewzta, judges Vitaly Aksenov Aksenov239, Nikolay Budin Budalnik, Dmitry Filippov DimaPhil, Borys Minaiev qwerty787788, Ilya Peresadin pva701, Grigory Shovkoplyas GShark, Artem Vasiliev vartem, Nikolay Vedernikov Niko, Ilya Zban izban.

And now the short tutorial. To get more details, you can read the programs of the judges solutions, published at the official site together with the official tests.

A. Closing ceremony

Probably the easiest way to solve the problem is greedy. Sort people from the first line by increasing of their stamina. Give them tickets in this order, each time using the place which is furthest away from the other line. After that try to assign people from the second line to the remaining seats by sorting people by stamina and seats by the distance.

The time complexity of your solution must not exceed O((nm)2), however using std::set one can get a solution with complexity of O(nmlog(nm)).

B. Cactusophobia

Let us divide the graph to biconnected blocks. Each block is either a bridge, or a cycle. Our goal is to remove one edge from each cycle, so that the number of remaining colors were maximum possible.

Let us build a bipartite graph, one part would be blocks, another one would be colors. For each block put an edge of capacity 1 for each color of an edge in this block (make multiple edges, or bigger capacity if there are several edges of some color). Add two vertices: source and sink, add edges from source to blocks, if the block is a cycle of length l, set its capacity to l - 1, if it is a bridge, set its capacity to 1. Add edges from color vertices to the sink of capacity 1.

It is quite clear that size of the maximum flow in this graph is indeed the answer to the problem.

As a final note, the judges know the solution that runs in O(n) and requires no maximum flow algorithms, challenge yourself to come up with it!

C. Homework

The solution is constructive.

  • First let us use backtracking to find solutions for all n, m < 5, it is also better to precalculate all answers, in order not to mess with non-asymptotic optimizations.
  • Now m ≥ 5.
  • Let us put asterisks from left to right, one row after another. When adding the new asterisk, we add 4 new L-trominoes, except the first asterisk in a row that adds 1 new L-tromino, and the last asterisk in a row adds 3 new L-trominoes. Let us stop when the number of remaining L-trominoes k is less than 4, or there are less than 5 free squares in the table.
  • Now there are two cases
    • there is a free row
    • thee is no free row
  • If there is a free row, we stopped because k is now less then 4. So:
    • k = 0: the solution is found
    • k = 1: if there are already at least 2 asterisks in the current row, put the asterisk in the beginning of the next row, if there is only 1, put it in the end of the current row
    • k = 2, k = 3 — similar, left as an exercise.
  • If there are now free rows left, one can see that you can only add k from the set {0, 1, 2, 3, 4, 5, 6, 8, 9, 12, 15} L-trominoes.
  • And finally there is also the special case where the size of the board is 3 × m, m ≥ 5 and k = 2 * (m - 1) - 8 — in this case the first column should be left empty, and the rest of the board must be completely filled with asterisks.
D. Slalom

First let us consider all paths from the starting square to the finish one. Let us say that two paths are equivalent, if each obstacle is at the same side for both paths. For each class of equivalence let us choose the representative path — the one that tries to go as low as possible, lexicographically minimum.

Let us use dynamic programming. For each square let us count the number of representative paths that go from the starting square to this one. When the obstacle starts, some paths can now separate. The new representatives will pass this obstacle from above (it will be to the right of them). So we add the sum of values for squares below it, but above any other lower obstacle, to the value for the square right above the obstacle.

To overcome the time and memory limits that the naive solution with O(nm) memory and O(nm2) time complexity, we use segment tree for range sum queries with mass update, running scanline and events "start of an obstacle", "end of an obstacle". This leads to the solution with O(m) memory and O(nlogm) time complexity.

E. Cipher

First let us consider a slow solution. Let us find a condition for each number k after what number of seconds Borya can distinguish it from the given number. Let us look at their initial encoding. Increase both numbers by 1 until the encodings are different (or one of the numbers needs more than n digits to represent in which case the beep allows Borya to distinguish the numbers as well).

Now there are two main ideas that allow us to get a better solution. First, we don't have to check all numbers. We only need to check numbers that differ from the given number in exactly one digit.

Second: to get the time when the numbers can be distinguished we don't need to iterate over all possible values. We just need to try all digit positions and all values for that position, and check only moments when the digit at the position will first have this value in one of the numbers.

So the complexity is now polynomial in n and since n ≤ 18, it easily fits into the time limit.

F. Array Covering

We give the outline of the solution, leaving technical details as an excercise.

First note that the answer will always use min(k - n, 0) subarrays with maximal sums. Sof let us find the sum of min(k - n, 0) maximal subarrays, elements that are used in them, and the following min(k, n) subarrays. We can do it using binary search for the border sum, and a data structure similar to Fenwick tree. For the given value of the sum this tree must provide the number of subarrays with greater or equal sum, sum of their sums, and the set of the elements of these subarrays. It should also allow to list all these subarrays in linear time complexity. This part has time complexity O(nlog2n).

Let us now describe how to solve the problem in O(n2logn). Let us try all values of x — the number of subarrays with maximum sums that we will use in our solution (there are O(n) variants, because top min(k - n, 0) subarrays will definitely be used). Let elements with indices i1, ..., im be the ones that are not used in these subarrays. Now we must add k - x segments that would contain all of these elements. Note that each of these k - x segments must contain at least one of these elements, and no two segments can have a common element among them (in the other case the solution is not optimal). These observations let us greedily choose these k - x segments in O(nlogn) and the final solution complexity is O(n2logn)

To optimize this solution, we keep segments [ij + 1; ij + 1 - 1] in the ordered set. When we iterate over x and increase its value, we remove some of the ij-s, and recalculate the required values for the affected segments. After that we must take k - x maximal values from the set, so since there are O(n) changes in total, this part now works in O(nlogn).

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By RussianCodeCup, history, 22 months ago, translation, In English,

Hi, everyone!

Tomorrow, on September 18, 2016, at 12-00 the Final Round of Russian Code Cup 2016 will take place. Top 50 participants of the Elimination Round will take part in the contest and have a chance to win great prizes. The length of the Final Round this year is 2 hours, the Final Round participants will compete online. You can watch the Final Round at

Meanwile we are pleased to announce that the Russian Code Cup team together with Codeforces project have prepared a small surprise to all those who didn't manage to get to the Final Round of RCC-2016. After the Final Round ends, at 14-05 Codeforces will host online-contest with RCC-2016 Final Round problems.

The contest will use ACM ICPC rules and will not be rated. Problems will be in English and in Russian. Problems were prepared for Russian Code Cup Final Round, so they are quite difficult, the contest is mostly suitable for Div1 participants. Of course, we would like to ask the Final Round participants do not use the contest for testing their upsolved solutions, please wait until the end of the online contest and use the problem archive.

So, everyone is welcome to watch the Final Round and then to take part in the online-contest! Good luck to everybody!

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By RussianCodeCup, history, 2 years ago, translation, In English,
A. Important Test

For each variant consider all prefixes of tasks. It is optimal to copy the task that takes most time, so we keep track of maximal time and calculate the sum S of tij of the prefix. If the maximal time is greater than t0 it is better to copy this task, getting the total time needed be S - max(0, M - t0). Now compare this value to t to check whether it is possible to write down that many tasks for this variant.

B. Centipede

First sort all notes based on the number of left legs. Then iterate over this array. If the centipede has li left legs, then first i notes can be correct, and other notes are definitely incorrect (we can ignore the fact that some notes have the same li, we will find the answer considering them).

Now we want to find out how many of the first i notes can be correct. Note with number j ≤ i can be correct if rj ≤ n - li. This problem is quite standard, we can solve it using, for example, Internal Tree.

C. Binary Tree

To minimize the number of days Basil needs, we need to minimize the height of the resulting tree. If the depth of the leaf in the resulting tree is h, Basil needs 2h - 1 - n days to create the required number of vertices.

Note that if the tree has a vertex of degree greater than 3, it is impossible to fulfill the task, because in a complete binary tree every vertex has at most 3 neighbors. Also note that the root vertex must have two sons.

So let us try all vertices with degree not greater than 2 and try it as a root. Choose vertex which results in the smallest tree height.

Finding the most distant vertex for each vertex in a tree is a well known trick. Choose vertex 0 as a root and run dfs(0). First find h[v] for each vertex — the length of the longest path down starting from v. Then run dfs2(0) passing the longest path up[v] to a vertex v through its parent. To find it when running dfs2(x) from y we choose maximum among up[y] + 1, and maximal value of h[z] + 2 for all children z of y except x.

D. Containers and Reagents

The solution proceeds in several rounds.

  • If sum(mini) > sum(vi) — the answer is «NO»;
  • If sum(maxi) < sum(vi) — the answer is «NO»;
  • Pour mini·pi / 100 of ci to each container;
  • For each reagent i sort containers that have cj = i by increasing pj and pour the rest of reagent i to them;
  • Pour (maxj - minjpj / 100 of reagent cj to the container j;
  • The rest of reagents can be put to containers arbitrarily;
  • If after doing so there is still extra liquid the answer is «NO»;
  • Some containers may still have not enough liquid.
  • Consider such container i, first pour from other container reagents other than this container is marked by, not making it contain less then minj of liquid
  • If it is still not enough, we can now move liquid cj from container j, it will not ruin percentage condition any more.

Now the resulting distribution of reagents satisfies all required conditions.

Final note: there could be precision troubles in this problem. Although there were no such tests, they are possible, Petr Mitrichev (congratulations on winning the round!) found such test after the contest. We understand that this is not good, and will try to avoid such problems in future.

E. Money Exchange

Let us first solve the following problem: find the minimal amount of money that cannot be paid by a set of coins. Let us first consider that we have no coins and the maximal amount we can pay is 0.

Let us describe the step when we have considered all coins with the value not exceeding X and we can pay any sum up to Y, inclusive. Let the total cost of coins with value greater than X, but not exceeding Y + 1, be sum. Then if sum = 0, we cannot represent Y + 1 with this set. In the other case we can move to the state where we have considered coins with the value not exceeding Y + 1, and we can pay any sum up to Y + sum, inclusive.

Now we see, that the value of the greatest considered coin grows as Fibonacci numbers, so the process terminates in at most O(log Answer) iterations.

So what we need to do now is to find the sum of numbers not exceeding X at a segment. This can be done using Interval Tree, the final complexity is O(m log n log Answer).

F. Elimination Round, Problem F

The answer is «NO» only if d is not divisible by the greatest common divisor of all ai. In any other case the answer exists. First let us create any array that satisfies a1·x1 + a2·x2 + ... + an·xn = d, but not necessarily other conditions. To do this divide all ai and d to gcd(a1, ..., an). If n = 1, then x1 = d / a1. In other cases find the corresponding values for any prefix [1, p] such values xi, p that a1x1, p + ... + apxp, p = gcd(a1, ..., ap). Use induction. Set x1, 1 = 1. If we have xi, p and want to find xi, p + 1. Use extended Euclid algorithm to find s and t: s·gcd(a1, ..., ap) + t·ap + 1 = gcd(gcd(a1, ..., ap), ap + 1) = gcd(a1, ..., ap + 1). Then for i ≤ p xi, p + 1 = s·xi, p and xp + 1, p + 1 = t. Getting xi, n, find xi = d / gcd(a1, ..., anxi, n = d·xi, n. This solution takes O(n2), and doesn't finish in time limit. To decrease the time complexity, first find subset of ai that has gcd equal to 1. Iterate over ai and add the current ai to the set if it decreases the number of different divisors. Such subset would contain at most 7 elements. Use the algorithm described above for this set, for other elements set xi = 0.

Now the algorithm is fast, but the conditions for xi are not satisfied: they can exceed 106 by their absolute value and there can be zeroes. Now let us normalize the values. First let us decrease xi to not exceed 106. To do this make xi non-negative, and less then a1 for all i > 1. This can be done by the following operation: subtract kai from x1 and add ka1 to xi where k = (xi mod a1 - xi) / a1. Note that the results of this operation can be stored in 64-bit integers, because x1 cannot exceed |d - a2x2 - ... - anxn| < 106·106·105. Now consider xi starting from the second one, and for each one do the following: subtract a1 from xi and add ai to x1. Note that there exists i, such that after applying the operation to x1, ..., xi we have 0 ≥ x1 ≥  - 106. Now we have |xi| ≤ 106, so everything that is left is to get rid of zeroes. Let us pair all zeroes to each other, except probably one if the number of zeroes is odd. For each such pair (i, j) set xi = aj and xj =  - ai. We probably have one index p left such that xp = 0. Consider several cases:

  • If there exists xj such that it is not equal to ap by its absolute value, then apply the following operation: subtract sign(xj)ap from xj and add sign(xj)aj to xp.
  • If there is no such xj, but ap ≤ 5·105, for any xj we can apply the following operation: add sign(xj)ap to xj and subtract sign(xj)ap from xp.
  • Finally if none of the above is true there must be aq, such that aq ≠ ap. Now do the following: subtract sign(xj)ap from xq and add sign(xj)aq to xp. Now xq is zero. But if n > 2, it is the first case for q, if n = 2 it is the second case for q.

Finally note that we must make this normalization must be performed for each prefix in the algorithm described in the first paragraph, so that the intermediate results fit 64-bit type. See judges solution for more details.

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By RussianCodeCup, history, 2 years ago, translation, In English,


Let us remind you that Russian Code Cup 2016 Elimination Round will take place on June 19, 2016 at 14-00 Moscow time. Top 200 from each qualification round can take part in Elimination round, 200 best coders in Elimination round will get branded championship t-shirt, and top 50 will advance to the Final Round that will take place in September. There are money prizes to grab in the Final Round.

Good luck everyone and see you at !

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By RussianCodeCup, history, 2 years ago, translation, In English,
A. Rectangle and Squares

To solve this problem, notice that the sides of the Elijah's rectangle don't matter, only the area is considered. So he can use any number of squares to create a rectangle of size C × nC for any positive integer n.

The optimal n is the result of the division AB / C2 rounded either up or down. So we just check two variants and choose the better one. Additionally, it is important to remember that n > 0, so sometimes it is impossible to round down.

B. Zeroes and Ones

First notice, that we can make inversions only in one string. Also notice that the order of inversions doesn't matter.

So the solution is greedy: consider characters of the first string from left to right. If the corresponding character s1[i] ≠ s2[i], we inverse characters i, i + 1 of the first string. In the end check that s1[n] = s2[n], if it's not, there is no solution, in the other case we have found the only, and therefore optimal, solution.

C. New Track

The solution is using special construction.

First, let us show how to create the track with maximal possible k. Starting with the horizontal segment, each vertical segment would cross all previous horizontal segments except the adjacent one. The example with 14 segments is shown in the picture.

To get exactly k segments first draw 2l segments such that l(l - 1) / 2 ≥ k > (l - 1)(l - 2) / 2, make a small track with k intersections. To do so, start creating the maximal example, but terminate the last segment after required number of intersections. The remaining n - 2l segments can be added in the beginning of the track to form the spiral around the smaller track. See picture below for example, where n = 17 and k = 9.

The constraint on k is really the maximal bound on the number of intersections of n segments. We omit the proof, but the hint is: look at the number of intersections of the pair of segments n and n - 1 with pairs n - 3 and n - 4, n - 5 and n - 6, ..., 2 and 1 (or only the segment 1, if n is even).

D. Tree

Let us add weights to tree edges, and set edge weight equal to 1 initially for all edges. Now instead of removing the vertex we would change the weight of the edge from it to its parent to 0. The distance between any pair of vertices would be equal to the distance that would be should we had actually removed the vertex.

We can solve the new problem as follows. For each vertex in the initial vertex find the distance from the root to the vertex, and put the distances to the array in DFS order. After changing some edge weight, we must change the distances in the continuous segment of the array. This can be implemented using, for example, range tree.

Now the distance between vertices can be found, using : dab = da + db - 2·dlca, where lca — is the least common ancestor of a and b, dv — is the distance from the root to v.

E. Barbarians

Notice that the angriness of all people in one connected component is equal to their initial angriness multiplied by some value, the same for all islands in this component.

Let us process edge removal in time proportional to the size of the smaller components that appears after its removal. In this case the time complexity of the algorithm would be . To prove the complexity notice that for each vertex when it is in the smaller component, the size of the component at least halves. So it can only happen times.

If removing the edge we could know which component is smaller, we could traverse it and move its vertices to a new component, leaving the vertices from the larger one in the old one. But we don't know which component is smaller. So let us run two BFS-s simultaneously in both components, and when one of them terminates, terminate the other one as well. This way the time for such BFS would be proportional to the size of the smaller component.

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By RussianCodeCup, history, 2 years ago, translation, In English,

Hi, all!

The last possibility to get to the Elimination Round of Russian Code Cup this year is the Third Qualification Round, that will take place on Sunday, June 5, 16:00. Everyone who has not yet qualified is welcome to participate. Good luck, and see you at Russian Code Cup 2016!

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By RussianCodeCup, history, 2 years ago, translation, In English,

Hi all!

The second qualification round of Russian Code Cup 2016 will take place on Sunday, May 29, 2016, at 12-00! We invite everyone who has not qualified yet to take part. Those who would not qualify in the second round are invited to take part in the third round on June 5. Good luck to everyone and see you at the round, don't forget to register for the championship at if you haven't done so already.

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By RussianCodeCup, 2 years ago, translation, In English,

A. Binary String

First notice that there is no solution in one of the following two cases:

  • Either |b - c| > 1
  • Or b = 0, c = 0, and both a ≠ 0 and d ≠ 0

For other cases there are two steps in the solution. First, construct the minimal string that satisfies conditions for 01 and 10 pairs. After that add a - 1 zeroes after the first zero, and d - 1 ones after the first one.

B. Train in a Tunnel

The solution is greedy.

First let's turn the light on in the first and in the last car. Then divide the train to one or more segments of continuous cars with light off. Сonsider one such segment. Iterate over cars and when you the sum of lengths of such cars would exceed h if we didn't turn the light on in the current car. Then turn the light in that car on, and continue from the next car.

C. Beautiful Partition

Consider a[1], it is either in M1, or in M2. Since it doesn't matter, let it be in M1. Then a[1] divides gcd(M1), therefore gcd(M1) — is the divisor of a[1].

Consider all divisors of a[1] (there are at most 1344 for numbers not exceeding 109). For each each divisor d all elements of the array that are divisible by dcan be put to M1 (in this case gcd(M1) would not be less then d, and the fewer elements are there in M2 — the greater gcd(M2) is), all the other elements can be put to M2. The answer can be relaxed with the value min(gcd(M1), gcd(M2)). For the purpose of relaxation we can ignore that M2 is empty and consider gcd for an empty set be equal to infinity. If we put any element to M2 the value of gcd(M2) will not be less then d in this case since all elements are divisible by d.

The time estimation is O(sqrt(a[1] + d(a[1])·n) where d(a[1]) ≤ 1344.

D. Problem Preparation

First let us solve the problem if there are no changes in ti. Let us generate new array cnt[j] equal to the count of i such that ti = j. Also calculate the array of prefix sums for cnt. Then for each k we can find the time for each friend in O(MAX / k) where MAX is the maximal time needed for one problem. We just iterate over all j and find the number of problems that require exactly j minutes for each friend using prefix sums array. The sum for all valid k of O(MAX / k) is O(MAXlogMAX).

Now let us see what happens if we change the time needed for the problem from t to t + 1. Note that only values for k which are divisors of t change. Since the number of divisors is small, we can just iterate over all divisors and change the answer just for them.

Similar idea works when t changes to t - 1, the answer changes only for such k that are divisors of t - 1.

E. Similar Subways

You have to find isomorphic connected subtrees of the two given trees with maximal number of vertices. Let us use dynamic programming to solve the problem.

Consider two directed edges (u1, v1) in the first tree and (u2, v2) in the second tree. Let the value dp[u1][v1][u2][v2] be equal to the maximal size of isomorphic subtrees, such that u1 is corresponding to u2 and vertices v1 and v2 are not included into the corresponding subtrees. Additionally, let us also calculate such value for v1 =  - 1 or v2 =  - 1, which means that any vertex from the first/second tree can be in a subtree. Then the answer to the problem is maximum over all u1, u2 of the values dp[u1][-1][u2][-1].

To calculate dp[u1][v1][u2][v2] we have to make correspondence between subtrees of u1 in the first tree and u2 in the second tree. Let us note that if e1 is the child of u1 not equal to v1, subtree of e1 contains fewer vertices then subtree of u1. So if we find the values in increasing order of sizes of subtrees, we can use values for all child subtrees. The value dp[e1][u1][e2][u2] gives us the maximal number we can get if we put correspondence between e1 and e2. So we can get the matrix a[e1][e2] which contains maximal values for each pair of adjacent vertices to u1 and u2, correspondingly. Now we have to find the maximal weight matching in this matrix which can be done by mincost flow, or Hungarian algorithm.

The complexity is O(n5).

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By RussianCodeCup, history, 2 years ago, translation, In English,

Hello, all!

We are glad to announce the final version of Russian Code Cup 2016 rules! The most important news of this year is that Russian Code Cup goes international, now the problems are available in both Russian and English. Anybody can participate, to enter the competition, please register at, participants of previous championships need to confirm their participation in their profile.

This year the prize pool is again 750 000 rub. We have changed the structure of the prizes to give more money prizes away, so top 25 now get cash. The winner gets 150 000 rub, the second place gets 100 000 rub, the third place gets 65 000 rub. Those who take places from 4 to 10 get 30 000 rub, and places from 11 to 25 can account for 15 000 rub. Top 200 in elimination round get branded Russian Code Cup t-shirts.

There are three levels of the championship. First you must qualify in one of the three Qualification Rounds (May 8, May 29 and June 5). Top 200 from each round proceed to Elimination Round (June 19), top 50 from it proceed to Final Round (September 18). If you fail to qualify in a Qualification Round you can still try in the following Qualification Rounds.

We invite you to the Qualification Round 1 on Sunday, May 8, 19-00 Moscow Time, and wish everyone good luck!

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By RussianCodeCup, 2 years ago, translation, In English,

A. Secret Code

All you have to do is just implement the required counting. Identifying correct characters at correct positions is absolutely straightforward. Identifying correct characters at incorrect positions can be, for example, implemented using used[c] array with true values for characters in the secret code and false for other characters.

B. Chaos

Consider two maximal numbers of the initial array. The maximal possible value that can be obtained is their average value.

To prove this, you can see Doc's moves as removing one number from the board and replacing two other numbers with two copies of their average value. So the average value of two maximal numbers never increases.

The following strategy allows to get such value in the end: each move remove the minimal number and replace two maximal numbers with two copies of their average values. Starting from the second move the two maximal numbers will be the same and equal to the answer.

C. New Adventure of Marty and Doc

You have to minimize the sum Sum(i = 1..n, j = 1..m, 2·ai, j·(|i - x| + |j - y| + 1)). First notice that no term depends on both coordinates, so the sum can be divided to three sums: Sum(2·ai, j), Sum(2·ai, j·|i - x|) and Sum(2·ai, j·|j - y|).

The first sum is constant for the given input, so we need to minimize the second and the third sum which can be done independently by choosing the row and the column. Consider the second sum, it can be proved that the optimal value is achieved if we take r equal to the median of Sum(j = 1..m, a1, j) times 1, Sum(j = 1..m, a2, j) times 2, ..., Sum(j = 1..m, an, j) times n.

Column can be found using the same idea.

D. Teams Creation

Let's use dynamic programming.

Sort all skill levels, and for each value find the number of students with such skill level. Note that if the team contains students with skill levels l and r, all students with skill levels m such that l < m < r will also be in the same team. So let us calculate the value dp[i][j][f] — the number of ways to create j teams of students with skill levels up to i, the last team can (f = true) or cannot (f = false) have more students. To make a transition we need additional value f[k][g] counts the number of ways to create g teams of k students that have the same skill level. Let there be k students with skill level i + 1, we iterate over g from 1 to k and relax dp[i + 1][j'][f'] using f[k][g]. Note that this takes O(nk) because the total number of g attempted for each j is equal to n.

To find f[k][g] we consider the k-th student. He is either in the team on his own (f[k - 1][g - 1]) or in one of the other teams (f[k - 1][gg).

Both arrays are calculated in time O(nk), so the time complexity is O(nk).

E. Maximal Sum

Let us sort bj in decreasing order. Add values to array a in decreasing order as well until all numbers greater or equal to the current bj is added. Now the array contains only numbers ai ≥ bj. After adding the new ai use interval tree to find the maximal sum subsegment at the segment (see discussion here, for example, to learn how to do this using interval tree:

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By RussianCodeCup, history, 2 years ago, translation, In English,


We are pleased to announce that in 2016 Russian Code Cup goes international! Problems will be available in two languages: Russian and English, everyone is welcome to participate!

Championship rules and time table will be announced next week, meanwhile we would like to invite you to take part in Warm Up round on Saturday, April 23, 18-00 Moscow Time (UTC+3). Register at and participate!

Good luck and see you at Russian Code Cup 2016!

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By RussianCodeCup, history, 3 years ago, In Russian,

Задача А. Сгибание ленточки.

Идея: Владимир Смыкалов.
Реализация: Дмитрий Филиппов.
Разбор: Дмитрий Филиппов.

В задаче дана полоска 1×2n, которую сначала n раз сложили ровно пополам, а потом развернули обратно. Складывания возможны двух способов: левую половину наложить на правую или правую половину наложить на левую. Требуется отвечать на запросы — по номеру сгиба узнать, ориентирован он вверх или вниз.

Научимся отвечать на запрос за O(log 2n), то есть за O(n). Суммарное количество запросов не превосходит 105, поэтому этого вполне достаточно. Будем эмулировать процесс сгибания для каждого запроса. Зная текущие длину ленточки и номер сгиба, легко понять, в какой половине этот номер находится, а затем, зная, в какую сторону произойдет сгиб пополам, можно найти и новое положение сгиба в укороченной в два раза ленточке. Для ответа на запрос одновременно с положением поддерживаем, в какую сторону сейчас ориентирован сгиб. Итого, получаем решение за O(qn), где q — суммарное количество запросов.

Задача B. Сбор монет.

Идея: Виталий Аксенов.
Реализация: Борис Минаев.
Разбор: Борис Минаев.

В задаче дана лента, разбитая на клетки, по которой ходит игрок. Каждую секунду в каждой клетке дополнительно появляется некоторое число монет, фиксированное для каждой клетки. Игрок за секунду может перейти в соседнюю клетку или остаться в текущей. Каждый раз когда игрок находится в некоторой клетке, он собирает все монеты, которые в ней находятся. Необходимо посчитать, какое максимальное число монет может собрать игрок за t секунд.

Заметим, что для каждой клетки важно знать только последний момент, когда игрок в ней находился. Рассмотрим путь игрока с конца. В каждый момент времени про некоторый непрерывный отрезок клеток уже известен последний момент их посещения, а для остальных клеток нет. Поэтому можно воспользоваться методом динамического программирования, состоянием которого будет отрезок посещенных клеток и текущее время. Кроме того необходимо хранить текущую позицию игрока. Заметим, что можно хранить только те позиции, когда игрок стоит на одном из концов отрезка посещенных клеток. Из каждого состояния существует не более чем два перехода — игрок может посетить клетку слева или справа от уже посещенных. Таким образом, время работы алгоритма составляет O(n2t).

Задача C. Топологическая сортировка и дети.

Идея: Артем Васильев.
Реализация: Виталий Аксёнов.
Разбор: Виталий Аксёнов.

В задаче дан граф и его топологическая сортировка, некоторые элементы которой были стёрты. Нужно корректно её восстановить топологическую сортировку.

Рассмотрим сначала алгоритм, а потом докажем его корректность. Вершины, для которых дан порядок, будем называть помеченными, остальные — непомеченными. Мы знаем какие числа не использовались в топологической сортировке, будем выставлять их по одному от больших к меньшим в соответствие вершинам. У нас есть очередное нерасставленное число. Первым делом удалим все уже помеченные стоки. Далее для каждого непомеченного стока узнаем максимальное значение в помеченной вершине, достижимой по обратным рёбрам. Для каждой вершины это число можно предподсчитать заранее: находим какую-либо топологическую сортировку графа и решаем задачу динамического программирования. В качестве соответствующей вершины нерасставленному числу выбираем любой сток, у которого предподсчитанное число наибольшее, и удаляем его из графа.

Предподсчёт чисел выполняется за O(V+E). Каждую итерацию нужно выполнять как можно быстрее, поэтому в каждой вершине мы храним количество оставшихся из неё исходящих рёбер. Когда мы удаляем сток, у всех вершин, из которых есть в него ребро, уменьшается исходящая степень, и если эта степень становится нулём, помещаем эту вершину в сет по предподсчитанному значению. Тем самым в нашем сете хранятся все стоки, и для выбора нужного нам достаточно взять самую последнюю вершину из сета. Каждая вершина кладётся в сет и вытаскивается ровно один раз. Итого время работы алгоритма: O(V·log(V) + E).

Теперь докажем корректность нашего алгоритма. Заметим, что нам достаточно доказать правильность первого шага алгоритма, далее используем метод математической индукции. Рассмотрим корректно заполненную топологическую сортировку p, предварительно выкинув все помеченные стоки. Далее рассмотрим сток s, который мы выбрали нашим алгоритмом для максимального числа, и сток, которому оно соответствует в топологической сортировке p. Если мы все числа в перестановке большие p(s) уменьшим на единицу, а выбранному нами стоку s выдадим максимальное число, перестановка останется корректной. Операция произошла корректно: мы сохранили свойство топологической сортировки для непомеченных вершин, а для помеченных вершин значение топологической сортировки не поменялось, так как по свойству выбора вершины s p(s) больше, чем значение топологической сортировки любой помеченной вершины.

Задача D. Правильный сад.

Идея: Артем Васильев.
Реализация: Артем Васильев.
Разбор: Артем Васильев.

В этой з[адаче дано множество точек на плоскости, и введено понятие хорошего множества точек. Хорошим множеством точек называется такое множество точек, что любой невырожденный прямоугольник со сторонами, параллельными осям координат и с противоположными углами в заданных точках содержит хотя бы одну другую точку из этого множества внутри или на границе. Требуется проверить, обладает ли заданный набор точек таким свойством.

Для начала докажем эквивалентное свойство: множество точек хорошее тогда и только тогда, когда для каждого угла такого прямоугольника существует точка на смежной с этим углом стороне. Очевидно, если выполняется это свойство, то выполняется и свойство, описанное в условии задачи. Однако, верно и следствие в противоположную сторону: рассмотрим произвольный прямоугольник с углами в точках A и B. По предположению, внутри или на границе этого прямоугольника существует другая точка из множества, назовем ее C. Если эта точка уже лежит на стороне, смежной с A, то наше утверждение верно. Иначе рассмотрим прямоугольник, построенный на точках A и C. Его площадь строго меньше, чем площадь исходного прямоугольника, а точек в множестве конечное число, значит когда-нибудь мы выберем точку, лежащую на стороне, смежной с A.

Сформулированное свойство помогает дойти до решения: для точки (xi, yi) обозначим за lefti такой максимальный x < xi, что в заданном наборе существует точка (x, yi). В случае, когда это значение не определено, считаем его равным равным минус бесконечности. Аналогичным образом введем величины righti, downi и upi. Рассмотрим область (lefti, righti) × (downi, upi). Если в этой области есть другая точка из заданного набора, то можно найти две точки, для которых построенный прямоугольник нарушает свойство. Стоит заметить, что подходит не любая точка из этой области. В качестве второй точки всегда подходит, к примеру, ближайшая по евклидовому или манхэттенскому расстоянию. Когда в этой области находится только точка (xi, yi), все прямоугольники содержат другую точку на стороне, смежной с ней. Таким образом, решение свелось к проверке n прямоугольников на наличие более одной точки внутри.

Реализовать такую проверку можно с помощью любой структуры данных, которая позволяет считать количество точек в прямоугольнике на плоскости: двумерное дерево отрезков (онлайн решение) или оффлайн решение со сканирующей прямой и одномерным деревом отрезков. Такое решение можно реализовать за O(n log(n)).

Задача E. Междуречье.

Идея: Виталий Аксёнв.
Реализация: Илья Збань.
Разбор: Илья Збань.

В задаче даны n непересекающихся монотонных ломаных и выпуклый многоугольник. Нужно узнать, какое минимальное число раз нужно приложить многоугольник, чтобы каждая ломаная имела хотя бы одну общую точку с одним из приложенных многоугольников (на границе или внутри многоугольника).

Заметим, что поскольку многоугольник выпуклый, а ломаные монотонны и не пересекаются, верен следующий факт: если многоугольник пересекает ломаные i, k, то для любого j, что i ≤ j ≤ k, этот многоугольник пересечет и j-ю ломаную. Используя этот факт, будем решать задачу жадно: найдем максимальное i1, такое, что существует многоугольник, пересекающий все ломаные с 1-й по i1-ю, добавим его к ответу, и продолжим: найдем максимальное i2 такое, что существует многоугольник, покрывающий ломаные с i1+1-й по i2-ю, и так далее. Число многоугольников, которые пришлось использовать, и будет ответом.

Научимся для ломаной ik находить значение ik+1. Это максимальный индекс, такой, что существует многоугольник, покрывающий ik+1-ю и ik+1-ю ломаную. Решим сначала более простую задачу: есть точка A и отрезок BC, нужно проверить, существует ли многоугольник, содержащий внутри себя и данную точку, и какие-то точки отрезка. Для этого построим сумму Минковского этого многоугольника и его самого же, инвертированного относительно (0, 0) (то есть, с противоположными по знаку координатами всех точек). Это какой-то новый многоугольник, назовем его P, с O(n) вершин, содержащий внутри себя точку (0, 0). Сдвинем его так, чтобы точка (0, 0) перешла в точку A, и проверим, что отрезок BC пересекается со сдвинутым многоугольником P — несложно убедиться, что это необходимое и достаточное условие.

Научившись проверять, существует ли многоугольник, пересекающий точку и отрезок, проверить то же самое для двух ломаных совсем несложно: перебираем отрезок на одной ломаной, строим его сумму Минковского с многоугольником P, и проверяем, пересекается ли новый выпуклый многоугольник с каким-нибудь отрезком второй ломаной.

Пусть k — суммарное число вершин на всех ломаных. Найдём асимптотику используемых нами операций — O(n) на построение многоугольника P, O(n) на построение суммы Минковского P и отрезка ломаной и O(log n) на проверку пересечения выпуклого многоугольника и отрезка. Так как многоугольников нужно построить максимум k, а пересечь в худшем случае нужно каждый многоугольник с каждым отрезком, итоговая асимптотика времени работы равна O(nk + k2log n).

Задача F. Робот на дереве.

Идея: Борис Минаев.
Реализация: Борис Минаев.
Разбор: Борис Минаев.

В задаче дано неориентированное дерево из n ≤ 10 вершин. Про каждое ребро известна его прочность wi, которая не превышает 15. По дереву перемещается робот, который изначально находится в случайной вершине. Каждый раз робот равновероятно выбирает ребро из тех, по которым он может пройти, и перемещается по нему. Каждый раз когда робот проходит по ребру, его прочность уменьшается на единицу. Если прочность становится равна нулю, то ребро удаляется. Если у робота нет ребер, по которым он может пойти, он останавливается. Необходимо посчитать математическое ожидание длины пути робота.

Для начала решим более простую задачу. Предположим, что необходимо посчитать не математическое ожидание длины пути, а количество различных путей, по которым мог пройти робот. Представим, что мы зафиксировали вершины, в которых робот начал и закончил свой путь, а также количество раз, которые робот прошел по каждому ребру. Рассмотрим, какие существуют ограничения на эти количества. Во-первых, ребра, по которым робот прошел положительное число раз, должны образовывать связный подграф исходного дерева. Во-вторых, рассмотрим некоторую вершину и количество раз, которые робот прошел по всем ребрам, которые являются смежными с этой вершиной. Если эта вершина является концом пути, то каждое смежное с ней ребро должно использоваться ровно wi раз. Максимум два ребра смежных с вершиной должны быть использованы нечетное число раз. Причем, если эта вершина лежит на пути из стартовой в конечную, то ровно два ребра должны быть использованы нечетное число раз. Для стартовой и конечной вершины (если они не совпадают) должно быть ровно одно ребро, которое используется нечетное количество раз, либо ноль, если вершины совпадают. Для всех остальных вершин все смежные ребра должны быть использованы четное число раз.

Если все эти условия выполнены, научимся считать количество путей, которые удовлетворяют условию задачи и используют ребра фиксированное количество раз. Для каждой вершины рассмотрим все смежные с ней ребра. Мы знаем количество раз, которое робот прошел по каждому из них. Мы можем легко посчитать, сколько раз робот прошел по ребру в направлении от заданной вершины. В каком порядке робот мог проходить по этим ребрам? Во-первых, по ребру, которое лежит на пути к вершине, в которой робот закончил путешествие, робот должен пройти в последнюю очередь. По всем остальным ребрам он может идти в любом порядке. Подсчет количества таких способов является стандартной задачей. Чтобы посчитать общее количество путей, по которым может пройти робот, необходимо перемножить посчитанные значения для каждой вершины.

Заметим, что для подсчета общего количества путей можно воспользоваться методом динамического программирования. Посчитаем количество путей, которые проходят по некоторому ребру заданное число раз и каким-либо образом проходят по поддереву, которое ограничено этим ребром. При этом считается, что каждый раз, когда путь идет вверх по ребру, он сразу же возвращается назад по нему. Чтобы посчитать значение динамического программирования необходимо зафиксировать количество раз, которые робот прошел по каждому из ребер, которые идут в поддеревья, умножить на уже посчитанные значения для поддеревьев, а также на количество способов расположить переходы в поддеревья. Будем рассматривать поддеревья вершины по очереди и фиксировать количество раз, которые суммарно робот пойдет во все рассмотренные поддеревья. Рассматривая очередное поддерево переберем количество раз, которые робот перейдет в это поддерево. Домножим ответ на значение динамического программирования от поддерева, а также на количество способов расставить переходы в поддерево среди других переходов.

Вернемся к первоначальной задаче. Теперь вместо количества способов будем хранить вероятность выбрать такой путь, а также математическое ожидание его длины. Подсчет вероятности отличается от подсчета количества способов только тем, что при каждом переходе по ребру необходимо поделить значение на количество ребер, которые инцидентны данной вершине. Единственная проблема, которая возникает при этом заключается в том, что степень вершины не является константой, так как ребра удаляются во время путешествия робота. Чтобы посчитать значения динамического программирования для ребра воспользуемся следующей идеей. Зафиксируем все поддеревья, ребра в которые будут удалены, а также общее количество раз, которое робот будет идти из вершины по некоторому ребру. Будем восстанавливать путь с конца. Существует два возможных варианта, которые необходимо различать. Либо робот идет по ребру, которое не будет удалено, тогда нужно уменьшить количество ребер, по которым еще необходимо пойти роботу и перейти к меньшей задаче. Либо робот идет по ребру, которое будет удалено, тогда к общему количеству ребер, по которым роботу необходимо пройти, нужно добавить количество раз, которые он пройдет по данному ребру, а также увеличить количество существующих инцидентных ребер на единицу.

Всего существует O(2childrenwi) состояний в динамическом программировании для заданного ребра, и переход между ними осуществляется за O(1). Так как необходимо посчитать значение динамического программирования для каждого количества раз, в которых было использовано ребро, общее время работы алгоритма получается равным O(2n(∑wi)2).

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By RussianCodeCup, history, 3 years ago, In Russian,

Всем привет!

Финал Russian Code Cup в этом году пройдет онлайн, но финалисты могут по желанию приехать лично на одну из двух площадок на выбор: офис Mail.Ru Group в Москве или в Университет ИТМО в Санкт-Петербурге.

В то время как участники будут решать выданные задания, с 11:30 по московскому времени на сайте в прямом эфире будет транслироваться ток-шоу, посвящённое прошлому, настоящему и будущему программирования и высоких технологий.

Среди гостей такие известные люди как Николай Никифоров — министр связи и массовых коммуникаций РФ, Наталья Касперская — генеральный директор InfoWatch, Сергей Андреев — президент ABBYY, Болтунов Олег — член PostgreSQL Foundation, Алексей Пажитнов — изобретатель игры «Тетрис» и многие другие.

Гости будут обмениваться мнениями по самым разным вопросам, связанным с развитием IT и программирования. В качестве ведущих выступят Антон Комолов и Михаил Мирзаянов, руководитель Центра олимпиадной подготовки программистов Саратовского Государственного Университета.

Трансляция пройдёт 19 сентября с 11.30 до 16.00 московского времени на

Среди зрителей будут разыгываться призы, так что приглашаем всех присоединиться к трансляции Russian Code Cup, а финалистам желаем удачи!

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