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### pikmike's blog

By pikmike, history, 7 days ago, translation, ,

1327A - Sum of Odd Integers

Idea: vovuh

Tutorial
Solution (vovuh)

1327B - Princesses and Princes

Idea: BledDest

Tutorial
Solution (pikmike)

1327C - Game with Chips

Idea: Ne0n25

Tutorial
Solution (Ne0n25)

1327D - Infinite Path

Tutorial

1327E - Count The Blocks

Tutorial
Solution (Roms)

1327F - AND Segments

Idea: Ne0n25

Tutorial
Solution (Ne0n25)

1327G - Letters and Question Marks

Idea: BledDest

Tutorial
Solution (BledDest)

• +87

 » 7 days ago, # |   +78 D and E should have been swapped !
 » 7 days ago, # | ← Rev. 5 →   -11 Great Contest! Congratulations!
 » 7 days ago, # |   +37 In this contest Problem A is also very tricky.
•  » » 6 days ago, # ^ |   0 Yes! C++ solution for A had to handle int overflow with k*k.
•  » » 3 days ago, # ^ |   0 can someone please explain this in A : It is obviously greater than 2(k−1)−1 because k2≤n and it is obviously odd because the parity of n and k is the same.
•  » » » 3 days ago, # ^ |   0 Got It.
 » 7 days ago, # |   +20 For E i used a more dp like solution.The number of blocks of size n is allways 10. Let $bl(n)=10$$bl(n-1) = 2*10^2 - 10*2 = 180$$bl(n-2) = 3*10^3 - 180*2 - 10*3$$bl(n-3) = 4*10^4 - bl(n-2)*2 - 180*3 - 10*4$$bl(n-4) = 5*10^5 - bl(n-3)*2 - bl(n-2)*3 - bl(n-1)*4 - bl(n)*5$...This can be implemented in $O(n)$ with simple additions and multiplications. 74112302 (I use pow there, but that could be a multiplication, too)
•  » » 7 days ago, # ^ | ← Rev. 2 →   0 What's the logic behind this? I found this pattern but it didn't make sense to me. Thanks in advance.
•  » » » 7 days ago, # ^ |   +6 For say n=3 there are $3*10^3$ single digits. But some of those digits have neighbours which are same digit, we call them blocks.So, every digit not belonging to a block is a single digit. For n=3 this is $3000 - 2*180 - 3*10$ The 2 and 3 are the size of the blocks, 180 and 10 are the number of blocks.And that pattern repeats, since for any blocksize, if n is increased by one, the number of blocksize+1 is the same as before. Every block can be extended by one member.
•  » » » 7 days ago, # ^ |   +11 The reason for the pattern is that the total number of digits is always going to be n * 10^n, so you're just weighting each block count by its size and finding the value for the next n that makes it add up to the right value.
•  » » » » 5 days ago, # ^ |   0 Why total number of digits is always n*(10^n)? Shouldn't it be 10^n only? Thanks.
•  » » » » » 5 days ago, # ^ |   0 It is $10^n$ numbers, $n$ digits each.
•  » » 6 days ago, # ^ |   +1 RHS of bl(n-i) contains n-i terms hence a total computation of summation i over n which is quadratic. Is this what you mean?
•  » » » 6 days ago, # ^ | ← Rev. 3 →   +1 I solved this problem in the same way. You can maintain the value that you are subtracting off, and update as you go. if value[] is the value you want, psum[] is the partial sum of value[], and subs[] is the amount you are subtracting off, then the transitions are:subs[i] = subs[i+1] + psum[i+2] + 2*value[i+1]value[i] = (n-i+1) * 10^(n-i+1) — subs[i]psum[i] = psum[i+1] + value[i]There are other ways to do it, but this is how I did it in my code. 74286577
•  » » » » 6 days ago, # ^ |   0 I like this implementation.
•  » » 6 days ago, # ^ |   0 wow wow wow wow wow.....thanks man....#thankusomuch.
•  » » 6 days ago, # ^ |   0 How can you implement this in o(n)?Wont it take o(n^2)?
•  » » » 6 days ago, # ^ | ← Rev. 4 →   +4 I'm leaving the power part, just for to demonstrate how to do it in O(N) $bl(n - x) = bl(n - x + 1)*2 + bl(n - x + 2)*3 + ... + bl(n + x - x)*(x+1)$ $bl(n - (x + 1)) = bl(n - (x + 1) + 1)*2 + bl(n - (x + 1) + 2)*3 + ... + bl(n - (x + 1) + (x + 1)) *((x + 1)+1)$ The second equation can be written as. 2.$bl(n - (x + 1)) = bl(n - x)*2 + \left[bl(n - x + 1)*3 + ... + bl(n) *(x+2)\right]$And the first. $bl(n - x) = \left[bl(n - x + 1)*2 + bl(n - x + 2)*3 + ... + bl(n)*(x+1)\right]$ The difference is:$sum = bl(n - x + 1) + bl(n - x + 2) + ... + bl(n)$$bl(n - (x + 1)) = bl(n - x)*2 + sum + bl(n - x)$So we just have to update $sum$ each step.For me, this is bit annoying than the solution described in the tutorial
•  » » 5 days ago, # ^ |   0 Nice observation, since top-down is always easy to observe. Good job !
 » 7 days ago, # |   +4 Can anyone explain the solution of D with examples and the topic I must know before solving it?
•  » » 6 days ago, # ^ |   +5 You should read about Cyclic Permutations to understand this topic better.
•  » » » 6 days ago, # ^ |   +3 Ok thank you ^_^
 » 7 days ago, # |   +5 Can someone explain solution of E? The language in the editorial is very hard to follow for me.
•  » » 7 days ago, # ^ |   0 especially "blocks which first element is a first element of integer"
•  » » » 7 days ago, # ^ |   0 Exactly man
•  » » 7 days ago, # ^ | ← Rev. 2 →   +2 For E , consider we are finding answer for length k , let x be the number within block ( 0 <= x <=9) for say k=3 and n=7; then two cases are : 1. xxx++++ , ++++xxx 2. +xxx+++ , ++xxx++ , +++xxx+ these are ( n — k -1 ) For case a: u choose x in 10 ways , now immediate next to x has to be different because we are doing this for length k ,so it will be chosen in 9 ways , remaining we have ( n-k-1)positions left which can be filled with 10 numbers . For case b : u can understand similarly .
•  » » » 6 days ago, # ^ |   0 Ya this makes it clear what the editorial meant, thanks!
•  » » » 3 days ago, # ^ |   0 How can we be guaranteed that there are no collisions, like For n==5 and k==2 Case:1 00___ -> last two dashes have 10 chances each so 00_00 is possible. ___00 -> first two dashes have 10 chances each so 00_00 is possible. We are counting that twice write.but we are not supposed to be.
•  » » » » 3 days ago, # ^ |   0 yeah , i thought about that and this is the reason i could not solve in the contest , i too thought that there will be collision . But this is not a collision , we actually need to add these two . Just take some time , u will see . ( we need block count , focus on this ). See , for 00x** , counting the right x** gives u numbers which contains it and it implies this block was present in them . for , **x00 we do the same and this block was present in them . When we add these two values , we are adding blocks 00___ and ___00 and 2 different block because of their position . For eg , 00900 u need to do +2 and not +1 .
•  » » » » » 2 days ago, # ^ |   0 thanks man its nice to observe this thing For eg , 00900 u need to do +2 and not +1(block count). thanks
•  » » » 6 days ago, # ^ |   +3 Cool, Thanks! I understand the solution however it feels that at my current level it is very unlikely that i ll be able to come up with a solution like this :(
•  » » » » 6 days ago, # ^ | ← Rev. 2 →   0 Not at all. I'm sure you can do it with practice. I didn't get this during the contest either. But I had an improvement over the last few times when I had similar problems (when I was completely clueless). This time at least I was approximately right as I was close to a solution. The only difference between then and now has been doing some practice on paper. It will look a lot simpler when you do it on paper, than by reading and trying to comprehend. Use this text as reference and don't get overwhelmed by it.
•  » » » » » 6 days ago, # ^ |   0 I hope so man, thanks again.
•  » » » 3 days ago, # ^ |   0 How can we be guaranteed that there are no collisions, like For n==5 and k==2 Case:1 00___ -> last two dashes have 10 chances each so 00_00 is possible. ___00 -> first two dashes have 10 chances each so 00_00 is possible. We are counting that twice write.but we are not supposed to be.
•  » » » » 39 hours ago, # ^ | ← Rev. 3 →   0 The problem asks for blocks, not how many unique numbers match the criteria. A hint in that direction is the second example: when n = 2, and k = 1, you'd expect 90 numbers, but the answer is 180, since it is blocks. A block is simply the same digit repeated to a certain length. So the number 23 would have 2 blocks of size 1 in it, even though it is only one number.Note that if you were to calculate the count of unique numbers, instead of blocks, then you'd divide this by a certain factor (your example correctly identifies the overcounting that needs to be accounted for).
•  » » » » » 35 hours ago, # ^ |   0 thanks dude , little bit confused with blocks and numbers,finally understood.
 » 7 days ago, # |   +7 Could someone explain the time complexity for problem D?
•  » » 7 days ago, # ^ | ← Rev. 10 →   +19 $O(n^{1/3})$ is a "rule of thumb" approximation for $\max _{x=1} ^{n} d(x)$, the maximum number of divisors of $x \in [1,n]$. For exact values, you may consult this list.Theoretically, this formula is $O(n^\epsilon)$ for any $\epsilon > 0$, but it takes very large values of $n$ to converge, so this fact's not useful in typical competitive programming contexts.You can get a close approximation at typical CP-relevant ranges ($n \lesssim 10^{18}$) via the formula $\max _{x=1} ^{n} d(x) \approx \min (3.5273 n^{1/3}, n^{1.066 / \ln \ln n})$ (source)
•  » » » 5 days ago, # ^ |   0 can you explain the 3rd test case of problem D. im getting ans with p5 but how for p2Q8 7 4 5 6 1 8 3 2 5 3 6 4 7 5 8 4solution In the third test case, p2=[3,6,1,8,7,2,5,4] and the sequence starting from 4: 4,p2[4]=8,p2[8]=4,… is an infinite path since c4=c8=4.
•  » » » » 5 days ago, # ^ | ← Rev. 5 →   0 For each index, you need to build the cycle for that index (and mark the other members of the cycle in an auxiliary array so you don't build any cycle more than once). Then for each divisor $d$ of the cycle size $s$, you want to find a position $i$ where all indices $cycle[i + kd]$ for $k \in [0, \frac s d)$ have the same color. A success means you have found a path in $p^d$, so the smallest $d$ of all successes is the answer.Sample: 74200072
•  » » » » » 5 days ago, # ^ |   0 Thanks really a great explanation, it was very hard for me understand this without a perfect explanation.
 » 7 days ago, # | ← Rev. 2 →   0 For D, example 2 of the problem statement states:In the second test case, p5=[1,2,3,4,5] and it obviously contains several infinite paths.However, if you take the original array p1 = [2 3 4 5 1], and run through the iterations, then: p1 = [2 3 4 5 1] p2 = [3 4 5 1 2] p3 = [5 1 2 3 4] p4 = [4 5 1 2 3] p5 = [2 3 4 5 1] So is there an issue with the statement, or in the way p5 has been calculated?
•  » » 7 days ago, # ^ | ← Rev. 2 →   0 $p^k[i]$ isn't calculated from $p^{k-1} [p^{k-1} [i]]$It is $p^k[i] =p[p^{k-1}[i]]$It was mentioned in the problem that multiplications like c=a*b turns out to be like c[i] = b[a[i]] So $p^k=p^{k-1}*p$ turns out to be $p[p^{k-1}[i]]$
•  » » » 7 days ago, # ^ |   0 Clear now. Thank you very much.
•  » » » 6 days ago, # ^ |   0 In Problem D, if t=1 , n=3 , p[]=[ 2, 3, 3 ] , c[]=[ 2, 1, 3 ]. As we already have an infinite cycle at p[3] of length 2. So output must be 1. But above code of pikmike is giving 3. So, please point out my mistake, if any.
•  » » » » 6 days ago, # ^ |   0 elements of the permutation are unique. It's mentioned in the problem "The second line contains n integers p1,p2,…,pn (1≤pi≤n, pi≠pj for i≠j) — the permutation p."
•  » » » » » 6 days ago, # ^ |   0 Oh, Sorry I didnt notice it. Thanks for the reply
•  » » » » » » 6 days ago, # ^ |   0 no probs :)
•  » » » » 6 days ago, # ^ |   0 the array p[] is a permutation of 1 to n , it can't have 2 3 3
•  » » » » » 6 days ago, # ^ |   0 I should have noticed it :) Thanks for the reply
 » 7 days ago, # |   +17 I think this contest was prepared very well. Especially solutions of C and E were satisfying. Thanks to authors.
 » 7 days ago, # |   0 Can this editorial be linked to the contest? Thanks!
 » 6 days ago, # |   -10 Why is the time complexity for problem B O(n + m), what is m as mentioned in the editorial? Shouldn't it be O(n^2), because k can be any value up to n.
•  » » 6 days ago, # ^ |   0 Yes, k can be any value up to N but. You didn't read the guaranteed carefully as there is a line told that.It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5. So if you run through all K. it's only up to 10^5 for all test cases. You can think M as the total number of kingdoms in lists.
•  » » » 6 days ago, # ^ |   0 Oh! my mistake thank you for correcting me
 » 6 days ago, # |   0 PROBLEM A can someone explain the condition : if(n%2!=k%2) then "NO"thanks
•  » » 6 days ago, # ^ |   0 Odd+Odd=Even.Even+Odd=Odd.Therefore you can only make N from sums of K odd number if and only if they have the same parity.
•  » » » 6 days ago, # ^ |   0 Clear now. Thank you very much.
 » 6 days ago, # |   +38 For the problem G, there needs little effort to reduce the complexity from $O(L |S|)$ to $O(m L^2 + |S|)$. Who is interested can compare my two submissions 74205780 and 74200846.
•  » » 6 days ago, # ^ |   +2 stO 叉姐 Orz, can you explain the details? thanks a lot!
•  » » » 6 days ago, # ^ |   +30 Divide the string $s$ with questions marks into $(m + 1)$ parts $p_0, p_1, \dots, p_m$. Essentially, we need $\mathrm{go}(u, p_i)$ for each node $u$ in the Aho-Corasick Automaton and its corresponding contribution. Direct computation leads to $O(L |S|)$ as the editorial says. However, if $p_i > L$, all $\mathrm{go}(u, p_i)$ agree on the value. Thus, we need only compute $\mathrm{go}(u, p_i)$ for one $u$.
•  » » » » 6 days ago, # ^ |   0 wow, thank you! I just realized I used the same technique on a contest I prepared last year(
 » 6 days ago, # |   +3 The use of c both as a permutation and the colour function makes it confusing at times.
 » 6 days ago, # |   0 can anyone explain me how does we can simply apply down , up, left ,right, as normal. because in problem C they have different notatation if we want to change the position.
 » 6 days ago, # |   -8 In Problem D, if t=1 , n=3 , p[]=[ 2, 3, 3 ] , c[]=[ 2, 1, 3 ]. As we already have an infinite cycle at p[3] of length 2. So output must be 1. But above code of pikmike is giving 3. So, please point out my mistake, if any.
•  » » 6 days ago, # ^ |   0 The p[] array must consist of disinct elements .But in your test case 2nd and 3rd elements are both 3 .
•  » » » 6 days ago, # ^ |   -8 Oh sorry , i didnt notice that :) Thank for replying
 » 6 days ago, # |   0 Who can help me explain F clearly! Thanks!
•  » » 6 days ago, # ^ |   +1 See my comment
 » 6 days ago, # |   0 Can someone help me in the solution for problem D. I got the question but the editorial is hard to follow. Thanks in advance!
•  » » 6 days ago, # ^ |   +2 The D question demands lot of observations and tricks . Read The editorial line by line just as hints and try them yourself . I assure that may help a lot
•  » » » 6 days ago, # ^ |   0 I did it over and over, still not able to catch up. Can you help me by providing some relevant prerequisites or a dry run of the solution?
 » 6 days ago, # |   0 I can't understand the DP defination of problem F. Would anyone explain for it clearly? Thanks.
•  » » 6 days ago, # ^ | ← Rev. 11 →   +27 First note that we can calculate a result for each bit separately and multiply them all for the final answer. Iterate $s$ over $[0, k)$, and let $b_i$ represent the bit in position $s$ of $x_i$.We can now divide the conditions into what we'll call $1$-segments and $0$-segments in accordance to its $b_i$ for the current $s$.Note that most of our arrays and data structures we use should be indexed over $[0, n+1]$ (we'll see why in a moment). The $[l, r]$ in the conditions should remain 1-indexed.Now we do the precalculation step. We want two things from the precalculation: For each index, we want to know whether it is covered by a $1$-segment. This is a classic "union of segments" problem and can be done in many ways: pre-sorting and two pointers, removal from sorted sets, or even DSU For each index $i$, we want to know the maximum $l$ among all $0$-segments whose $r < i$. If you're stuck, see the spoiler for details on how to do this efficiently. We'll call these values $f_i$ in accordance with the editorial. (If no such segments exist for index $i$, $f_i = 0$.) SpoilerFirst initialize all $f_i$ to $0$. For each $0$-segment, set $f_{r+1} \leftarrow \max (f_{r+1}, l)$. Then iterating over $[1, n+1]$, set $f_i \leftarrow \max (f_i, f_{i-1})$We can now perform the DP. $dp_i$ represents the number of arrays of length $i$ that end with a $0$, and all $0$-segments contained within it contain at least one $0$.$dp_0$ represents the empty array and therefore should be $1$.Now we iterate $i$ over $[1, n+1]$If $i$ is covered by a $1$-segment, $dp_i = 0$ as we can't place a $0$ in this position.Otherwise $\displaystyle dp_i = \sum_{j = f_i}^{i-1} dp_j$ because the last-seen zero cannot be in a position earlier than $f_i$. This can be calculated efficiently by maintaining a prefix sum over $dp$.The result for bit $s$ is equal to $dp_{n+1}$. This is because we can always fix a $0$ at the fictitious $n+1$-th index, and delete it to get a suitable array.Asymptotics $\tilde{O}(k(n + m))$ with a possible $\log$ factor somewhere depending on implementation. Bonus: Solve in strict $O(k(n + m))$.sample: 74254156 (uses a specialized DSU-like data structure for the union-of-segments problem)
•  » » » 5 days ago, # ^ |   0 What about the segments which are neither in 0-segment nor in 1-segment?
•  » » » » 5 days ago, # ^ |   0 It's fine. Because you've already precalculated $f_i$, if there are no constraints on the index, $f$ would not increase and the summation will continue accumulating.
•  » » » 5 days ago, # ^ |   0 Thank you so much!
•  » » » 2 hours ago, # ^ | ← Rev. 2 →   0 I'm finding it bit hard to understand why $dp_{i}=\sum_{j=f_{i}}^{i-1}dp_{j}$. Can you please explain.
•  » » » » 5 minutes ago, # ^ | ← Rev. 2 →   0 $dp_j$ represents the set of strings that are length $j$ and end with a zero. So for each of those strings, you are able to add ones until index $i-1$, and then add a zero to form part of $dp_i$'s set.
 » 6 days ago, # |   0 For D, I can see why it holds that on raising by power m it divides into gcd(m,L) cycles (L=lenth of original cycle) but why does it hold when m does not divide L? Can someone give intuition/mathematical reason for this? I can see that it holds but why?
•  » » 6 days ago, # ^ |   0 When we increase the power by say 2, we are traversing the original cycle by skipping 2 steps at a time. For example: for a cycle 1 -> 2 > 3 -> 4 -> 1, with k = 2, we would have 1 -> 3 -> 1 as the cycle if we take 1 as start. Now we would have gcd(k, L) unique such cycles, in this case 2 cycles (other one being 2 -> 4 -> 2).
•  » » 6 days ago, # ^ | ← Rev. 6 →   0 when gcd(m,L) = 1 L*x%m = 0 only when x == m,essentially you have to walk through every single element in the cycle until you return to c1 because all mutiples of L mod m should be distinct else assume x and y are integers < m and x>y then if Lx%m = Ly%m,then (x-y)L%m = 0 which is a contradiction since gcd(L,m) = 1 and x-y < m
 » 6 days ago, # | ← Rev. 2 →   +1 I spent some time looking at D and wanted to improve upon the given solution with external resources. Notice that P is nothing more than a cyclic permutation. Lecture notes on Permutation Groups and Symmetric groups. Each permutation can be written in Disjoint cycle notation. Thus there exists cycles. let A = (a1 a2 a3 .. am) be a cycle. then A^k can be written in Disjoint cycle notation with gcd(k, m) cycles each of cycle length m / gcd(k, m). Proof So let us look at all cycles A = (a1 a2 ... am). We need to take a look at all subcycles of length x. This corresponds to every cycle when we raise A to k. since x = gcd(k, m) from 2, x | m. Thus we can skip a lot of values of x. If a particular answer of x has the property that all colors in it are same then ans = min (ans, x). Why min(ans, x) ans not min(ans, k)? 1. We don't know exact k that it would have been raised to. 2. Notice gcd(k1, m) == gcd(k2, m) => same cycle[properties of cyclic group] Thus x is the least value of k such that gcd(m, k) = x as x | m.
 » 6 days ago, # |   -14 Wow，我想到这种方法了，但是把2nm当做2(n+m)了，It's a pity
 » 6 days ago, # |   0 Can anybody explain problem A.I couldn't understand the editorial.
•  » » 6 days ago, # ^ |   +1 So basically , there are only 2 cases in which the output can be "No" , ones where the parity of n and k is different because odd number of odd numbers can only add upto an odd number and similarly for 2n odd numbers .And the second case where we get NO as a result will be when the sum of first k odd numbers exceeds n. Understand this as if the minimum of the k odd numbers (that is the first k odd numbers) add up to be more than k , then there is no way to choose k numbers such that there sum=n.example: 7 3 here ist 3 odd numbers , i.e 1+3+5 = 9 , so there is no way we can hava a solution for this case.Also it is important that the k odd numbers should be distinct.
•  » » » 5 days ago, # ^ |   0 Thanks
 » 6 days ago, # |   0 Can someone please help with a detailed code for Q2.
 » 5 days ago, # |   0 can someone explain me how in A, that parity being equal and k^2<=n be the only two conditions we need to check?
•  » » 5 days ago, # ^ |   0 sum of odd numbers when their quantity is even leads to even no while when quantity is odd it leads to odd no. eg- 1+3=4(qty=2 and sum is even), 1+3+5=9(qty=3(odd) and sum is odd)sum of first k odd natural number is k*k. So min sum is k*k. n should always be greater than k*k
 » 5 days ago, # |   0 I am still not able to understand the implementation of problem E .Please someone explain me in a easier way.
 » 5 days ago, # | ← Rev. 2 →   0 In Q1327B can there be more than one ans?(for test case 2) eg- IMPROVED 1 1 or IMPROVED 2 1
 » 5 days ago, # |   0 For Problem D : Can someone explain this in statement in detail ? "Now, walking with step k we can note, that the initial cycle c split up on GCD(k,m) cycles of length m/GCD(k,m)"
•  » » 5 days ago, # ^ |   0 think an array which can be divided into x group of equal size of s. If we transition from first element in first group for s time, it will become first element of second group, similarly first element of group 2 will become first element of group 3 by transitioning it s times and so on. If all visited element have same color, it will be an infinite cycle. This is what we are doing at below line, finding one infinite cycle in editorial solution.for(int pos = s; pos + step < sz(cycle); pos += step)Now we will find all number which can divide our cycle size of equal group and traverse as above to find the answer.
 » 5 days ago, # |   0 In the editerial for problem D, "for k1 and k2 such that GCD(k1,m)=GCD(k2,m) the produced cycles will have the same sets of vertices and differ only in the order of walking", I can vaguely understand it, but who can prove it rigorously? why "the same sets of vertices"?
•  » » 5 days ago, # ^ |   0 Suppose, you have the cycle $c_0, c_1, \dots, c_{m-1}$. Since $p^k[c_i] = c_{(i + k) \mod m}$, then $c_i$ and $c_j$ will be on the same split up cycle iff $i + kx \equiv j \mod m$ or $i + kx - my = j$ or $j - i = kx - my$.There is a property that all solutions of $kx \pm my$ with an integer $x$ and $y$ is divisible by $GCD(k, m)$, so it means that $i - j$ is divisible by $GCD(k, m)$.Finally, since $GCD(k_1, m) = GCD(k_2, m)$ then it's obvious that the same pairs of indices will be on the same split up cycles.
 » 5 days ago, # |   0 Please help me with Problem E: We choose remaining digits as 10(n-len-1)(in the first case), but if in case those remaining digits also contain a block of size len — doesn't it mean we count same blocks more than once(in 2nd case again)?? For example, if n = 7 len = 2, the numbers like 1121134 come in both 1st and 2nd case?
•  » » 5 days ago, # ^ |   0 as you see, the numbers 1121134 has two blocks of length 2, which is exactly the sum of 1st and 2nd case.
•  » » » 5 days ago, # ^ |   0 got it, thanks a lot!
 » 5 days ago, # |   0 Problem E: Why don't we count the same blocks a couple of times in the first case? When we multiply 10 * 9 * 10^(n — len — 1), there are some cases included when the right side is also a block with the same length, and then we count that twice by multiplying 2.
 » 5 days ago, # | ← Rev. 3 →   0 Can someone help me with a TLE on problem G? I'm struggling to determine if my time complexity is the problem or if I just need to reduce my constant factor. It looks like a lot of the other solutions are using bfs to compute the subproblems (I'm using recursion) but I'm not sure what the benefit of that approach is. Here's my submission 74512969.edit: problem was that I was using LONG_MIN instead of LLONG_MIN...
 » 4 days ago, # |   -8 I am not sure if i am not understanding the question but i think for input 132 3 21 2 3answer should be 2 but the solution provided gives answer 3. Plz can any one explain me where if i am wrong because i think 2 & 3 form a loop
 » 4 days ago, # |   0 can somebody tell me what's wrong with my approach for problem C? link :- https://codeforces.com/contest/1327/submission/74527919 thanks in advance!!
•  » » 46 hours ago, # ^ |   +4 Here is my analysis (had to read multiple comments and the editorial to arrive at this).Now ($p^k[c_i] = c_{(i+k)mod m}$) :this is similar to jumping k steps at a time. Now each sub-cycle will be produced by starting at i (for simplicity taking ($c_i$) as i), and then after jumping k steps for x no. of times we will reach back at i. This completes our sub-cycle i.e. ($i + kx \equiv i mod m$) where x as you can see is the no. of jumps taken to reach back at i, also is the length of this sub-cycle.Or, ($i + kx - my = i \implies kx = my$) for some integer y.Now to find the min. x such that above equation holds, you see k and m are fixed, so kx = my = lcm(k, m). And as we know lcm(k, m) = km/gcd(k, m) = km/g.Therefore, ($x = lcm(k, m)/k \implies x = km/kg \implies x = m/g$). Hence proved length, x of sub-cycle produced is size/gcd(size, k). Now to prove that no. of subcycle will be gcd(k, m):you can see that each sub-cycle will of equal size, starting with some i, j, ... , also each element has to be a part of exactly 1 such subcycle, thus product of no. of subcycles and size of each such subcycle must be 'm', hence the proof follows.For proof of second observation as per editorial, please refer adedalic comment.Hope this helps :)
 » 10 hours ago, # |   0 [problem:[problem:1327B — Princesses and Princes]]Timeout is coming in C++. Same code Test#6 `#include #include #include #include #include #include #include #include #include //max/min //lower_bound/upper_bound #include //ceil/floor using namespace std; /////////////////////////////////////////////////////////////////////// int main() { // freopen("input.txt", "r", stdin); // freopen("output.txt", "w", stdout); int T; cin >> T; for(int t=1;t<=T;t++) { int N; cin>>N; vector> arr; arr.resize(N); vector marriedP(N); vector marriedD(N); for(int i=0;i>k; arr[i].resize(k); for(int j=0;j>arr[i][j]; if(marriedD[i] == 0 && marriedP[arr[i][j]-1]==0) { marriedP[arr[i][j]-1] = 1; marriedD[i] = 1; } } if(improve == 0 && marriedD[i] == 0) { improve = 1; a = i; } } if(improve==1) { for(int j=0;j