传送门:http://codeforces.com/contest/1092/problem/D1 D1. Great Vova Wall (Version 1) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vova's family is building the Great Vova Wall (named by Vo…

传送门:http://codeforces.com/contest/1092/problem/D2 D2. Great Vova Wall (Version 2) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vova's family is building the Great Vova Wall (named by Vo…

Codeforces Round #527 (Div. 3) 题解 题目总链接:https://codeforces.com/contest/1092 A. Uniform String 题意: 输入n,k,n表示字符串的长度,k表示从1-k的小写字符(1即是a),现在要求最大化最少字符的数量. 题解: 贪心搞一搞就行了. 代码如下: #include <bits/stdc++.h> using namespace std; int T; int n,k; int main(){ cin>…

一场div3... 由于不计rating,所以打的比较浪,zhy直接开了个小号来掉分,于是他AK做出来了许多神仙题,但是在每一个程序里都是这么写的: 但是..sbzhy每题交了两次,第一遍都是对的,结果就涨了.. A - Uniform String 没什么意思.. #include<cstdio> #include<cstring> #include<algorithm> #include<queue> #include<set> #inclu…

Codeforces Round #529 (Div. 3) 题目传送门 题意: 给你由左右括号组成的字符串,问你有多少处括号翻转过来是合法的序列 思路: 这么考虑: 如果是左括号 1)整个序列左括号个数比右括号多 2 2)在这个位置之前,所有位置的前缀左括号个数都不少于前缀右括号个数 3)在这个位置和这个位置之后,在修改后所有位置的前缀左括号个数减去前缀右括号个数大于2 (这里这么想,把左变成右,左-1,右+1) 右括号也是这样 代码: #include<bits/stdc++.h> usi…

http://codeforces.com/contest/1092/problem/D1 Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches. The cur…

D1. Magic Powder - 1 题目连接: http://www.codeforces.com/contest/670/problem/D1 Description This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraints, then you can just write a single soluti…

任意门:http://codeforces.com/contest/1118/problem/D1 D1. Coffee and Coursework (Easy version) time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The only difference between easy and hard versions…

链接:https://codeforces.com/contest/1130/problem/D1 题意: 给n个车站练成圈,给m个糖果,在车站上,要被运往某个位置,每到一个车站只能装一个糖果. 求从每个位置开车的最小的时间. 思路: vector记录每个位置运送完拥有糖果的时间消耗,为糖果数-1 * n 加上消耗最少时间的糖果. 对每个起点进行运算,取所有点中的最大值. 代码: #include <bits/stdc++.h> using namespace std; typedef lon…

D1. RGB Substring (easy version) time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output The only difference between easy and hard versions is the size of the input. You are given a string s consistin…