z Given two integers $$$n$$$ and $$$b$$$ (in decimal notation), your task is to calculate the number of trailing zero digits in the $$$b$$$-ary (in the base/radix of $$$b$$$) representation of $$$n\,!$$$ (factorial of $$$n$$$). a1an Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. The problem statement has recently been changed. This is the first time I get FST :D I struggled to understand the problem and missed that case, Like Frozen said,It means that i passed the pretests during the contest, but i failed the main tests after the contests and my answer is not accepted. Thanks for the creators of such an amazing contest! I didn't manage to solve B but for me it's the best problem I've ever seen on codeforces. The problem statement has recently been changed. k 2n2 In the second step: the array is $$$[0,1,2]$$$, so the minimum non-zero element is 1. You can multiply it by two $$$15$$$ times. The subsegment is some consecutive segment of elements of the array. GitHub: Let's build from here GitHub In the first step: the array is $$$[1,2,3]$$$, so the minimum non-zero element is 1. Example 1: Input: matrix = [ [1,1,1], [1,0,1], [1,1,1]] Output: [ [1,0,1], [0,0,0], [1,0,1]] Example 2: There is only one subsegment with the sum $$$0$$$. Then $$$t$$$ test cases follow. WLOG, assume that $$$x_1 \leq x_2 \leq x_3 \leq x_4$$$. Now, notice: Hence, we have the approach of the editorial. a1ayaz Hey, I had the same idea but didn't get it to work during the contest. The only programming contests Web 2.0 platform. to check, returns that value. = ) k But I felt B was a lot harder than normal, My solution to D within $$$2n-5$$$ queries: 145449958. ) 2*n-2 There's a typo in 2 of your if conditions, the third and fourth one. You have to help Kolya and change his array in such a way that it doesn't contain any subsegments with the sum $$$0$$$. If you've seen these problems, a virtual contest is not for you - solve these problems in the archive. When we perform standard range additions, the change in $$$x_n - x_{n-1}$$$ is zero for points in the middle of the array; in this case, we see that for points in the middle of the update, the change in $$$x_n - x_{n-1} - x_{n-2}$$$ is zero. For each test case, print the answer the minimum number of moves required to obtain such an array that each its element is divisible by $$$k$$$. It is . 2). 4 My solution to D in around $$$1.5n$$$ queries 145469547, Video editorial for anyone wanting looking. It seems to be more difficult than the official solution :-(. A array a[5]={1,1,0,2,2}.If you iterate over all n-2 possibilities for the last number,you will find that all result are the same ,but 0 isn't one of the first 2 numbers. For problem D: Finding Zeroes: For every 4 queries that the editorial makes, you can actually do it in 3. m n z I checked it once more, the fact i.e there is one and only one zero leads my code to run correctly. For me, B is one of the best Div2 problems in 2022 so far. Smallest number with at least n trailing zeroes in factorial. i That is, you have to tell us two numbers $$$i$$$ and $$$j$$$ and you win if $$$a_i = 0$$$ or $$$a_j = 0$$$. Example: 6 = 2^2 + 2^1 so now cnt[2]+=1, cnt[1] += 1 (the index is the power) and save the powers for the number (6) for step 2. y 0 Kolya got an integer array $$$a_1, a_2, \dots, a_n$$$. 2], I think I just did something crazy? Yes, for each row, connect (i,2*j) and (i,2*j+1), for 0<=j for each distinct value, connect the 2*i th position and the 2*i+1 th position of this value's occurrences. (a_{4}a_{n}) It is supported only ICPC mode for virtual contests. How do I set my home country on Codeforces? In one operation, you can: You are given $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$. It's so simple when you know the solution. If problem B was as follows : Alice starts with x and Bob starts with x+1 it woulda been much more solvable coz then one would find easily an answer to the question "What does x and x+1 mean or contribute to the solution" I don't see why the authors chose x+3 over x+1 maybe I missed something that makes it crucial to the problem. Avoiding Zero time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given an array of n n integers a1,a2, ,an a 1, a 2, , a n. You have to create an array of n n integers b1,b2, ,bn b 1, b 2, , b n such that: You needlessly use 10 modulo operations by a non-constant modulo per query, which is pretty damn slow. Once you have the index that produces the maximum result the second number or the last number is definitely the 0. Problem F's Time Limit is 1s,to be honest,I think it is meaningless,boring and FUCKING.My solution is $$$O(n)$$$ but I used "cin" to read a char.During the last 10 sec I submitted it.Then I got TLE on 7? I am not claiming index at min to be min and max to be max but instead saying that one of them is for sure min and max. The second line contains $$$n$$$ space-separated integers $$$a_1, a_2, \ldots, a_n$$$ $$$(1 \le a_i \le 10^9)$$$, the elements of the array. After printing a query, don't forget to output line feed and flush the output buffer. November 14, 2020 19:29. The second line of each test case consists of n n integers a1,a2, ,an a 1, a 2, , a n ( 1 ai n 1 a i n) the elements of the array a a. a n x 1 + Div. Which makes me curious, what was the motivation behind the time constraints in F? For this, consider the first 4 numbers to be 1, 1, 2, 2. If you've seen these problems, a virtual contest is not for you - solve these problems in the archive. 4. B and C probably should've been flipped but that says nothing about the quality of B. I think it was a great problem. a1a2az We obtained the required array. k By parina , 12 years ago , Getting WA on testcase 8. testcase 8 is. This is a very simple problem and strongly recommended to try this problem yourself before jumping to a solution directly. Otherwise you will get the verdict Idleness limit exceeded. The problem statement has recently been changed. 2), Codeforces Round 887 (Div 1, Div 2) Tutorial, 2022-2023 Southern And Volga Russian Regional - Editorial. And if it is not then it is going to be captured in our further iteration. 3 An efficient solution is to do following. 2 n2 Note that the array in each test case is fixed beforehand and will not change during the game. n Once you know the number of zeros, it is easier to find them. The only line of the input contains two integers $$$n$$$ and $$$b$$$ ($$$1 \le n \le 10^{18}$$$, $$$2 \le b \le 10^{12}$$$). You need to count the number of zero stretches, not the number of zeros. Then it's just inclusion-exclusion: we count the number of pairs of integers, minus the number of pairs with 1 bit in common, plus the pairs of integers with 2 bits in common, minus the pairs of integers with 3 bits in common Can you explain this in little more detail ? ax==0ay==0 , a Just increase $$$x$$$; $$$x=1$$$, $$$a = [1, 2, 1, 3]$$$. The solution works fine but its complexity is exponential For any number n, if its binary representation contains the k 0's then time complexity of above code is O(2k log2n). a If the current element is 1, then change it to 0 and increment the count by 1 and search for the 0 for the next operation as all consecutive 1's will be automatically converted to 0. a My motivation: the difficult part of this problem is clearly performing range Fibonacci updates, so our first order of business is to simplify the queries. count how many times the pattern appears in the trie -> 2^k (can be optimized to N maybe?) Does the same rule of thumb hold for other $$$k > 1$$$? constructive algorithms interactive math *2000, 2085. For problem C, I think the simpler and crucial observation is that any A.P, as long as the common difference is even, is divisible by n (no. 2^k = 10^6 which is acceptable, you can optimize it storing how many leafs are in each subtree, when your value has no more '? 0 0 buying bunch of higher number cost less per grape and that's why we will buy it first. Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. 1 + Div. a_{1}a_{2}a_{3}, ( count pairs with bitwise And zeros in array. Virtual contest is a way to take part in past contest, as close as possible to participation on time. or set $$$v = (2 \cdot v) \bmod 32768$$$. x j from equation 1 and 2 we have FirstElement & SecondElement = FirstElement. n z In addition, your solution uses absurd amount of extra % operations, that are very slow when modulo is not constant. Your profile picture is more than enough for distraction. Nevermind Fibonacci and suppose the operation was to increase the elements of a given subsegment by a given constant. Consider the first test case of the example: Note that you can't add $$$x$$$ to the same element more than once. If the second case, then the value of $$$(k, k+1)$$$ giving largest answer in this step must contain the zero. $$$a_1 = 19$$$. 2. Another approach for problem E without using eulerian circuit: If some number occurs odd number of times, it is obvious that no solution exists. Avoiding Zero Problem solution Codeforces Global Round 11- This Problem is taken from codeforces global round 11. Consider the first example. Number of Good Ways to Split a String. This is not going to work Take the following test case 3 3 6 0 for this the cnt Array will be like cnt[0] = 1, cnt[1] = 2, cnt[2] = 1 Now for 3, the answer is 3-cnt[0]-cnt[1] that is 0 But (3&0) == 0, so the above algorithm fails. The problem statement has recently been changed. In the first example, $$$6!_{(10)} = 720_{(10)} = 880_{(9)}$$$. Round 889 Question B, Interactive Problems: Guide for Participants, Atcoder problem statement of F Cans and Openers, UNIQUE VISION Programming Contest 2023 Summer(AtCoder Beginner Contest 312) Announcement. 0 The + and $$$\oplus$$$ operations have one thing in common: the parity of the res of A + B or A $$$\oplus$$$ B will be opposite with A only if B is an odd number. 0 2*n-2 0 , ningshancheng: If you consider $$$x_i$$$ as points on the number line, then each query gives you the length of the interval. 2 <= A <= 100000 Tree given in the form of an array B (one indexed) with A-1 elements where B [i] denotes parent of i+1 th node. 1 Im pretty sure it still works tho if u ignore the first thing I said about terminating early if last number query never changes. (Sorry for my poor English). 3

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