### drazil's blog

By drazil, history, 5 years ago, A. Numbers

If D > 0, then there are at most D - 1 integers between the X-th largest integer and the X-smallest integer. So the answer would be 2X + D - 1. If D = 0, then the X-th largest integer is the exactly same integer as the X-smallest integer. So the answer is 2X - 1. If D < 0, then the case with the largest number of integers would be: the X-smallest integer is the very next integer larger than the X-largest integer if sorted ascendingly. So the answer would be 2X - 2.

Overall, the answer can be expressed as 2X + max(D - 1,  - 2). Time complexity is O(1).

B. Power

There are at most 106 discrete time in the logs so if we can find out how many computers are in use at each time, we can calculate the final answer by summing the cost at each time in O(106) time. In order to find out that number, we could scan through the log and accumulate the number of on events and off events at each time which takes O(n) time. After that, just simulate those events time by time to get the number of computers in use at each time.

Time complexity is O(n + max(t)).

C. Robot and Maze

In a word, depth first search. But watch out for stepping into visited blocks to avoid looping. Moreover, you need to issue a command to go back to the previous block in the calling stack. You can use no more than answer × 4 moving commands to solve this task.

D. Combination

Firstly, you have to use inverse modular to deal with the division in the calculations. The combination function is straight forward but there are cases that the C(a + k, b + k) would become 0 for some intermediate values of k because of the modular operation. One way to handle this is to maintain the current C(a + k, b + k) using the form px × r where x is the largest possible integer. If x = 0 we know C(a + k, b + k) = r mod p, otherwise we know C(a + k, b + k) = 0 mod p.

Let's consider a scenario to illustrate the meaning of x. The task is to maintain the value n mod 3 with multiple and division operations. Initially n = 2 and the operations are 1.) multiply by 3 then 2.) divide by 3. If we just simulate each step, we would have n = 0 when the first operation is applied and never recover the true value after the second operation. So, with x, we would have x = 0, n = 2 initially. After the first operation we would have x = 1, n = 2. Since x ≠ 0 so we would know the answer is 0 now. After the second operation, we would have x = 0, n = 2 which preserves the truth that the answer is 2.

When maintaining C(a + k, b + k). There are two operations: multiple and division. To multiple by v, first we find x', r' such that v = px' × r', then do x = x + x', r = r × r'. To divide by v, first we find x', r' such that v = px' × r', then do x = x - x', r = r × modularinverse(r').

Time complexity is .

By drazil, 8 years ago, First I want to thank all the problem solvers!
Hope you had or are going to have a good time with Dreamoon!

If you think there's something that can be improved in this editorial please let me know!

Definitions used in this editorial:
⌈⌉ stands for ceiling function.
⌊⌋ stands for floor function. stands for real division.
For non-negative integer a and positive integer b, div(a, b) stands for integral division, .
For non-negative integer a and positive integer b, mod(a, b) stands for module operation, a = div(a, b) * b + mod(a, b).
length(string) is the length of string.
For non-negative integer a ≤ b, A[a..b] stands for the set of A[a], A[a + 1], A[a + 2]... A[b - 1] when A is an array and the substring of A consists of ath to (b - 1)th character(inclusive) of A when A is a string. For such substring we have length(A[a..b]) = b - a.
C(a, b) stands for the combination function, the ways of selecting a elements from a group of b elements.

476A - Dreamoon and Stairs
We can show that the maximum number of moves possible is n and minimal moves needed is , so the problem equals to determine the minimal integer that is a multiple of m in the range .
One way to find the minimal number which is a multiple of m and greater than or equal to a number x is , we can compare this number to the upper bound n to determine if there is a valid solution.
Although best practice is O(1), O(n) enumeration of each possible number of moves would also work.
time complexity: O(1)
sample code: 8212169
explanation of sample code:
The can be calculated in the following c++ code if a is non-negative and b is positive: (a+b-1)/b
Because / in c++ is integral division so (a+b-1)/b would result in Let a = div(a, b)b + mod(a, b) = db + m, .
Which means if , otherwise div(a + b - 1, b) = d + 1. Can be translated to if , otherwise div(a + b - 1, b) = div(a, b) + 1. Which matches the value of .

476B - Dreamoon and WiFi
The order of moves won't change the final position, so we can move all '?'s to the end of the string.
We have the following information:
1. the correct final position
2. the position that Dreamoon will be before all '?'s
3. the number of '?'s
We can infer that the distance and direction dreamoon still needs to move in the '?' part from 1. and 2., and furthur translate that to how many +1s and -1s dreamoon will need to move.
What's left is a combinatorial problem, the probability would be .
So we can compute that formula within O(n) time assuming n is the length of commands, but since N is small so we can brute force every possible choice of '?' with some recursive or dfs like search in O(2n) time complexity.
Note that the problem asks for a precision of 10 - 9, so one should output to 11 decimal places or more.
time complexity: O(n), assuming n is the length of commands.
sample code: 8215177

476C - Dreamoon and Sums / 477A - Dreamoon and Sums
If we fix the value of k, and let d = div(x, b), m = mod(x, b), we have :
d = mk
x = db + m
So we have x = mkb + m = (kb + 1) * m.
And we know m would be in range [1, b - 1] because it's a remainder and x is positive, so the sum of x of that fixed k would be .
Next we should notice that if an integer x is nice it can only be nice for a single particular k because a given x uniquely defines div(x, b) and mod(x, b).
Thus the final answer would be sum up for all individual k: which can be calculated in O(a) and will pass the time limit of 1.5 seconds.
Also the formula above can be expanded to . Dreamoon says he's too lazy to do this part, so if you use O(1) solution you just computed the answer faster than Dreamoon!!!
Note that no matter which approach one should be very careful of overflowing of the integer data type of the used language. For example one should do a module after every multiplication if using 64-bit integer type. And pay attention to precedence of operations: take c++ for example a+b%c would be executed as a+(b%c) instead of (a+b)%c, another c++ example a*(b*c)%m would be executed as (a*(b*c))%m instead of a*((b*c)%m).
Thanks saurabhsuniljain for pointing out the preceding problem and examples in the comment!
time complexity: O(1)
sample code: 8215188

476D - Dreamoon and Sets / 477B - Dreamoon and Sets
The first observation is that if we divide each number in a set by k, than the set would be rank 1. So we could find n sets of rank 1 then multiple every number by k.
For how to find n sets of rank 1, we can use {6a + 1, 6a + 2, 6a + 3, 6a + 5} as a valid rank 1 set and take a = 0 to n - 1 to form n sets and thus m = (6n - 1) * k.
The proof that m is minimal can be shown by the fact that we take three consecutive odd numbers in each set. If we take less odd numbers there will be more than 1 even number in a set which their gcd is obviously a multiple of 2. And if we take more odd numbers m would be larger.
The output method is straight forward. Overall time complexity is O(n).
time complexity: O(n)
sample code: 8215198

476E - Dreamoon and Strings / 477C - Dreamoon and Strings
First let A[i] to be the minimal length L needed so that substring s[i..i + L] can become pattern p by removing some characters. We can calculate this greedily by keep selecting next occurrence of characters in p in O(length(s)) time for a fixed i, so for all i requires O(length(s)2).
Next we can do a dp D[i][j] where D[i] is the answer array for s[0..i]. D[i] can contribute to D[i + 1] with 0 or 1 removal and to D[i + A[i]] with A[i] - length(p) removal(s). In other words, D[i][j] can transition into tree relations D[i + 1][j] = D[i][j] //not delete s[i], D[i + 1][j + 1] = D[i][j] //delete s[i], and D[i + A[i]][j + A[i] - length(p)] = D[i][j] + 1 //form a substring p by deleting A[i] - length(p) characters. Calculate forwardly from D to D[length(s) - 1] gives the final answer array as D[length(s)]. Calculating D[i] requires O(length(s)) time for a fixed i, so for all i takes O(length(s)2) time.
time complexity: O(n2), n = length(s)
sample code: 8215203

Another solution:
Let k = div(length(s), length(p)). We can run an edit distance like algorithm as following (omitting the details of initialization and boundary conditions):

for(i=0;i<n;i++)
for(j=0;j<k*p;j++)
if(s[i]==p[j%length(p)])
D[i][j] = D[i-1][j-1]
D[i][j] = min(D[i][j], D[i-1][j] + (j%length(p)!=length(p)-1))


That means remove cost is 1 when it is in the middle of a p and 0 elsewhere because p need to be consecutive(thus no need to be actually remove outside of a p). Then D[n][t * length(p)] is the minimal number of removals to have t non-overlapping substring of p. So we have answer[D[n][t * length(p)..(t + 1) * length(p)] = t. And after the maximal t is reached, decrease answer by 1 for every length(p).
time complexity: O(n2)
sample code: 8215394

477D - Dreamoon and Binary
Let Xb be the binary string of number X. An ideal sequence can be expressed as a partition of Xb: P1 = Xb[1..p1], P2 = Xb[p1..p2], ... PK = Xb[pK - 1..length(Xb)] where Pi ≤ Pi + 1. The length of operations of such sequence is PK + K.
We can calculate the number of different ideal sequences by dp. State D[i][j] stands for the answer of state that we have print Xb[1..j] and last partition is Xb[i..j]. A possible way of transition is that a state D[i][j] can go to state D[i][j + 1] and D[j][k] where k is the minimal possible position such that value of Xb[j..k] is equal to or greater than the value of Xb[i..j] and Xb[j] is 1 since we can't print any leading 0.
Note that D[j][k + 1] can also derived from D[i][j] but it will covered by D[i][j] → D[j][k] → D[j][k + 1], so we don't need to consider this case to avoid redundant counts.
If we can determine k for each i, j pair in O(1) then we can compute this dp in O(length(Xb)2) in the following manner:

for(j=0;j<n;j++)
for(i=0;i<j;i++)
compute the transitions of D[i][j]


So let's look into how to calculate the value k for a given i, j pair. If the value of Xb[j..2j - i] is equal to or greater than Xb[i..j] than k is 2j - i because if k is less than 2j - i would make length of the new partition less than the previous partition thus its value would be lesser. And if k can't be 2j - i, the value of 2j - i + 1 is always a valid choice because it would make the length of the new partition greater than the previous one. So for each length L if we know the order of Xb[i..i + L] and Xb[i + L..i + 2L] in O(1) time we can calculate k in O(1) time(can be easily shown by assuming L = j - i).
One way of doing such is using prefix doubling algorithm for suffix array construction to build a RMQ structure for query in O(1) time. The prefix doubling algorithm requires O(nlgn) precompute time. Note there is still a various of ways to do this part of task in the same or better time complexties.
And for the shortest length part we can compute the minimal parts needed so far for each state along with the preivous dp. Then compare all states ends with j = length(Xb).
Overall we can solve this problem in O(length(Xb)2) with caution in details like boundaries and module operations.
time complexity: O(n2), n = length(Xb)
Note the sample code use a nlgnlgn version of prefix doubling algorithm
sample code: 8215216

Although swapping two parts of a query would result in different answer, if we reverse the lines length alltogether then the answer would stay the same. So we only analyze queries where r2 ≥ r1.
The answers would comes in several types, we’ll discuss them one by one:
1. HOME key being pressed once: the answer will be 1(HOME) + r2 - r1(down keys) + c2(right keys). Note that this is the best answer if the HOME key is ever pressed once, so we won’t consider HOME key anymore. This step is O(1) for a single query, thus O(q) in total.
2. direct presses r2 - r1 down keys and no or one END key in those rows: because the cursor position will be reset to the row length if we go down to a row with length less than current cursor position, the possible positions that can be at row p if we start at end of each previous rows can be track with a stack. The position directly pressing only down keys from r1 to r2 is min(c1, the length of first row after r1 in stack). We can use a binary search to get the first row after or equal r1 in stack. From that row till r2 in the stack are the positions possible when pressing ONE END key (pressing more won’t get more possible positions), we can use a binary search to find the position closest to c2 which is the best. We can sort all queries and use O(qlgn) time for queries and O(n) time for maintaining stack, so O(qlgn + n) in total.
3. go back some rows and press one or no END key at the row: we only interested in END OF rows in stack constructed in 2.. We can use a binary search in stack to identify the range of rows interested. For those lengths of row less than c2(also can use a binary search to locate) we only need to consider the first one encountered(closest to r1), note still need to consider if it needs an END key for this case. For those lengths of row greater than or equal to c2, the answer would be r1 + r2 - 2 * (row reached) + (length of the row) - c2 + (1 if length of the row greater than c1). The terms related to row in the previous formula is 2 * (row reached) + (length of the row) + (1 if length of the row greater than c1). We can identify the range with and without +1(last term) and query in a segment tree of the minimal value of 2 * (row reached) + (length of the row) for each range. Thus the time need here is O(nlgn) for maintaining segment tree and O(qlgn) for query while sharing with the stack of 2., so O(nlgn + qlgn) in total.
4. go beyond r2 some rows and press no or one END at the row: this needs a reversed stack and very similar approach of 3.. The time complexity is the same as 2. and 3..
So the total time complexity is O((n + q)lgn), note this problem is very hard to get the code right.
time complexity: O((n + q)lgn)
sample code: 8212528 Tutorial of Codeforces Round #272 (Div. 2)
By drazil, 8 years ago, Hello, everyone! Codeforces Round #272 will be held at this local time. We're looking forward to your participation!

The problems are from dreamoon_love_AA and drazil(that's me) from Taiwan, and thanks 9mmlitswe for some discussion. Also we want to thank Zlobober and Gerald for helping us prepare the round, Delinur for translating the statement into Russian, and MikeMirzayanov for Codeforces and Polygon.

This is our first round on Codeforces, we hope you'll find it interesting! Please read all problem statements and discover what the main character dreamoon_love_AA do in those problems for he's really cute =)

Update1 Note this round will be held 1.5hrs earlier than usual Codeforces rounds, so please double check the starting time in your local time.

Update2 Score distribution!
Div2: 500-1500-1500-2000-2500
Div1: 500-1000-1500-2000-3000

Update3 The contests are over. Congratulations to the winners!

Div1:
1. Petr
2. qwer1561
3. kutengine
4. ifsmirnov
5. TankEngineer

Div2:
1. ridowan007
2. a00012025
3. xavier13540
4. v_Enhance
5. pkq2006

And standings are here:
Div1: standings
Div2: standings

Update4 Editorial can be found here. It's finished by now but it's welcome to tell me if anything can be improved! Announcement of Codeforces Round #272 (Div. 1)