time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given two integers n and k. Your task is to construct such a string ss of length nn that for each ii from 1to k there is at least one…
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…
Educational Codeforces Round 40 (Rated for Div. 2) C. Matrix Walk time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There is a matrix A of size x × y filled with integers. For every , *A**i, …
http://codeforces.com/contest/1092/problem/A You are given two integers nn and kk. Your task is to construct such a string ss of length nn that for each ii from 11 to kk there is at least one ii-th letter of the Latin alphabet in this string (the fir…
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…
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…
#include<bits/stdc++.h>using namespace std;const int maxn=1e6+7;pair<string,int>p[maxn];int nn,n;int cmp(pair<string,int>a,pair<string,int>b){    return a.first.length()<b.first.length();}char ans[maxn];multiset<string>sst…
题意:给你某个字符串的\(n-1\)个前缀和\(n-1\)个后缀,保证每个所给的前缀后缀长度从\([1,n-1]\)都有,问你所给的子串是前缀还是后缀. 题解:这题最关键的是那两个长度为\(n-1\)的子串,我们只要判断哪个是前缀就行了,然后再遍历一遍所给的子串,用长度为\(n-1\)的前缀子串来判断是子串是前缀还是后缀. 代码: int n; string s[N]; bool vis[N]; int cnt; int main() { ios::sync_with_stdio(false);…
#include<bits/stdc++.h>using namespace std;int a[200007];stack<int>s;int main(){    int n;    int mn=0;    scanf("%d",&n);    for(int i=1;i<=n;i++){        scanf("%d",&a[i]);        if(a[i]>mn)            mn=a…