Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmer…

B. A Trivial Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks f…

A Trivial Problem Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among thos…

B. A Trivial Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks f…

B. A Trivial Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks f…

//把一个序列转换成严格递增序列的最小花费 CF E - Sonya and Problem Wihtout a Legend //dp[i][j]:把第i个数转成第j小的数,最小花费 //此题与poj 3666相似 a[i]转换成a[i]-i #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #inclu…

B. A Trivial Problem 题目连接: http://www.codeforces.com/contest/633/problem/B Description Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such t…

Problem description Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among th…

数学技巧真有趣,看出规律就很简单了 wa 题意:给出数k  输出所有阶乘尾数有k个0的数 这题来来回回看了两三遍, 想的方法总觉得会T 后来想想  阶乘 emmm  1*2*3*4*5*6*7*8*9*10...*n 尾数的0只与5有关 是5的几倍就有几个0  因为5前面肯定有偶数 乘起来就有一个0 而且最后输出肯定是连续的5个 hhh 兴奋 开始上手 乱搞一下  发现复杂度还行 测样例 发现 k=5 的时候不对了 输出25~29了 应该是0的 咦  测了一下 25!应该是6个0的 25=5*5…

Problem - C - Codeforces 难度: 1300 input 6 aabbdabdccc zyx aaababbb aabbcc oaoaaaoo bmefbmuyw output 3 3 2 0 2 7 开始没调出来, 去看DE了, 就不该往后看, 但是当时思维没跳出来, 也不一定做出来, 呜呜当时是想着每次更新需要看前一次什么时候有这个字母此题重要一点更新就是, 越靠前进行配对的字母才是最优, 把问题转化一下, n-配对的字母数即为结果, 这里贴两个代码, 第一个是小小不像…