F:考虑对于每个字母对求出删掉哪些字符集会造成字符串不合法,只要考虑相邻出现的该字母对即可,显然这可以在O(np2)(或小常数O(np3))内求出.然后再对每个字符集判断是否能通过一步删除转移而来即可. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using nam…

Problem   Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship Time Limit: 2000 mSec Problem Description Input Output The only line should contain the minimal number of days required for the ship to reach the point (x2,y2)(x2,y2). If it…

Problem   Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems Time Limit: 3000 mSec Problem Description Input The input contains a single line consisting of 2 integers N and M (1≤N≤10^18, 2≤M≤100). Output Print one integer, the total n…

Educational Codeforces Round 60 (Rated for Div. 2) 题目链接:https://codeforces.com/contest/1117 A. Best Subsegment 题意: 给出n个数,选取一段区间[l,r],满足(al+...+ar)/(r-l+1)最大,这里l<=r,并且满足区间长度尽可能大. 题解: 因为l可以等于r,所以我们可以直接考虑最大值,因为题目要求,直接求连续的最大值的长度就是了. 代码如下: #include <bits…

Educational Codeforces Round 84 (Div. 2) 读题读题读题+脑筋急转弯 = =. A. Sum of Odd Integers 奇奇为奇,奇偶为偶,所以n,k奇偶性要相同. 由求和公式得k个不同奇数组成的最小数为k2,所以n≥k2. #include <bits/stdc++.h> using namespace std; void solve(){ int n,k; cin>>n>>k; if((n-k)%2==0&&…

A. Best Subsegment 题意 找 连续区间的平均值  满足最大情况下的最长长度 思路:就是看有几个连续的最大值 #include<bits/stdc++.h> using namespace std; ; int a[maxn]; int main(){ int n; scanf("%d",&n); ; ;i<n;i++)scanf("%d",&a[i]),maxnum=max(maxnum,a[i]); ; ; ;i…

#include <bits/stdc++.h>using namespace std;const long long mod = 1e9+7;unordered_map<long long,long long>mp;long long n,m;long long dp(long long n){    if(n<0)        return 0;    if(n<m)        return 1;    if(mp.find(n)!=mp.end())//已经…

#include <bits/stdc++.h>using namespace std;int main(){ string t; cin>>t; int n=t.size(); string s1(n,'a'),s2(n,'a'),s3(n,'a'); for(int i=0;i<n;i++){  s1[i]=char('a'+(i%26));//从a到z循环  s2[i]=char('a'+((i/26)%26));//第i位为(i/26)%26+'a',保证了26*26…

time limit per test 2 second memory limit per test 256 megabytes input standard inputoutput standard output You a captain of a ship. Initially you are standing in a point (x1,y1)(x1,y1) (obviously, all positions in the sea can be described by cartesi…

题目传送门 题意: 一个魔法水晶可以分裂成m个水晶,求放满n个水晶的方案数(mol1e9+7) 思路: 线性dp,dp[i]=dp[i]+dp[i-m]; 由于n到1e18,所以要用到矩阵快速幂优化 注意初始化 代码: #include<bits/stdc++.h> using namespace std; #define mod 1000000007 typedef long long ll; #define MAX 105 ;//矩阵的大小 int T; ll n,m; ll add(ll…