1) Link to the problem: http://codeforces.com/contest/888/problem/C 2) Description: You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant if each substring of s with length at least k contains this character…

http://codeforces.com/contest/888 A Local Extrema[水] [题意]:计算极值点个数 [分析]:除了第一个最后一个外,遇到极值点ans++,包括极大和极小 [代码]: #include<bits/stdc++.h> using namespace std; int main() { +]; int maxn,minn; maxn=minn=; cin>>n; ;i<=n;i++) { cin>>a[i]; } ;i&l…

Problem A Lucky Year 题目传送门[here] 题目大意是说,只有一个数字非零的数是幸运的,给出一个数,求下一个幸运的数是多少. 这个幸运的数不是最高位的数字都是零,于是只跟最高位有关,只保留最高位,然后加一求差就是答案. Code /** * Codeforces * Problem#808A * Accepted * Time:15ms * Memory:0k */ #include<iostream> #include<cstdio> #include<…

After several latest reforms many tourists are planning to visit Berland, and Berland people understood that it's an opportunity to earn money and changed their jobs to attract tourists. Petya, for example, left the IT corporation he had been working…

Vasya has an array a consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the seco…

Digital collectible card games have become very popular recently. So Vova decided to try one of these. Vova has n cards in his collection. Each of these cards is characterised by its power pi, magic number ci and level li. Vova wants to build a deck…

You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist. Divisor of n is any such natural number, that n can be divided by it without remainder. Input The first line contains two integers n and k (1 ≤ n ≤ …

A permutation p of size n is an array such that every integer from 1 to n occurs exactly once in this array. Let's call a permutation an almost identity permutation iff there exist at least n - k indices i (1 ≤ i ≤ n) such that pi = i. Your task is t…

题目链接 题意:给你两个数n,m,和一个大小为n的数组. 让你在数组找一些数使得这些数的和模m最大. 解法:考虑 dfs但是,数据范围不允许纯暴力,那考虑一下折半搜索,一个从头开始往中间搜,一个从后往中间搜.在中间相遇的时间二分更新最大值即可. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mp make_pair #define pb push_back usin…

题目链接:E. Maximum Subsequence 用了一个Meet-in-the-middle的技巧,还是第一次用到这个技巧,其实这个技巧和二分很像,主要是在dfs中,如果数量减小一半可以节约很多的时间.  Meet in the middle(有时候也叫作split and merge)是一种用以获取足够高效解决方案的灵巧的思想.和分治思想非常类似,它将问题分割成两个部分,然后试着合并这两个子问题的结果.好处在于通过使用一点额外的空间,你可以解决两倍规模的原来可以解决的问题. #in…