Bob is an avid fan of the video game "League of Leesins", and today he celebrates as the League of Leesins World Championship comes to an end! The tournament consisted of nn (n≥5n≥5) teams around the world. Before the tournament starts, Bob has…

Long is a huge fan of CFC (Codeforces Fried Chicken). But the price of CFC is increasing, so he decides to breed the chicken on his own farm. His farm is presented by a rectangle grid with rr rows and cc columns. Some of these cells contain rice, oth…

Hanh lives in a shared apartment. There are nn people (including Hanh) living there, each has a private fridge. nn fridges are secured by several steel chains. Each steel chain connects two different fridges and is protected by a digital lock. The ow…

Bob watches TV every day. He always sets the volume of his TV to bb. However, today he is angry to find out someone has changed the volume to aa. Of course, Bob has a remote control that can change the volume. There are six buttons (−5,−2,−1,+1,+2,+5…

把每一次输入的一组数字存下来,然后把每个数字出现的组数存下来 然后找只出现过一次的数字a,那么这个数字a不是开头就是结尾,默认为开头(是哪个都无所谓),然后去找和它出现在同一组的两个数字b和c,而b和c同时出现的只有两组,除了已经知道的,就是需要去查找的,然后循环,同时存在的组数,找到两个数字a和b同时出现的组而且不是第一组,找到之后的组的第三个数字,就是要接在后面的数字,然后循环这个操作 #include <bits/stdc++.h> using namespace std; ][];//…

Codeforces Round #601 (Div. 2) ---- 比赛传送门 周二晚因为身体不适鸽了,补题补题 A // http://codeforces.com/contest/1255/problem/A /* 签到题 简单的贪心题 本考虑过是否有先小于再加上去会更小的情况 但两种情况恰好步数是一样的 */ #include<iostream> #include<cstdio> #include<cmath> using namespace std; int…

传送门 A. Changing Volume 签到. Code /* * Author: heyuhhh * Created Time: 2019/11/19 22:37:33 */ #include <bits/stdc++.h> #define MP make_pair #define fi first #define se second #define sz(x) (int)(x).size() #define all(x) (x).begin(), (x).end() #define…

cf 上的一道好题:  首先发现能生成所有数字-N 判断奇偶 就行了,但想不出来,如何生成所有数字,解题报告 说是  所有数字的中最大的那个数/所有数字的最小公倍数,好像有道理:纪念纪念: #include<iostream> #include<stdio.h> #include<algorithm> #include<cstring> #include<algorithm> #include<cmath> using namespa…

D. Berland Fair time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output XXI Berland Annual Fair is coming really soon! Traditionally fair consists of nn booths, arranged in a circle. The booths ar…

Codeforces Round #601 (Div. 2) E2. Send Boxes to Alice (Hard Version) N个盒子,每个盒子有a[i]块巧克力,每次操作可以将盒子中的一块巧克力左移或右移,要求移动后的每个盒子中的巧克力数量都能被k整除(无视空盒子),求最小的操作数.(1<=N<=1e6,0<=a[i]<=1e6) 题解 k只需考虑巧克力总数的质因子 考虑每个盒子的贡献,盒子中可以保存k的整数倍(k*i)块巧克力,从左向右递推,每个盒子保留最大数目的…

#include <bits/stdc++.h> using namespace std; typedef long long ll; ; int a[N]; int n; bool prime(int x) {//判断是否为质数 ; i*i <= x; i++) { ) return false; } return true; } ll solve(int x) { vector<int>b; ll ans = ; ; i <= n; i++) { &&…

//为了连贯,采取一条路形式,从第一行开始 也就是s型 #include <bits/stdc++.h> using namespace std; ; char str[MAXN][MAXN]; vector<char> ch;//存放鸡的名字 void init() { '; i++) ch.emplace_back(i); for(char i='A'; i<='Z'; i++) ch.emplace_back(i); for(char i='a'; i<='z';…

//题目要求的是每一个点最少要有两条边连接,所以可以先构成一个环.然后再把剩余的最短的边连接起来 #include<iostream> #include<algorithm> using namespace std ; ; int n,m; struct edge { int num; int w; } a[N]; int T; bool cmp(edge a,edge b) { return a.w<b.w; } void solve() { ; ; i<=n; i+…

好吧,其实我拿到这个题的时候,首先想到了bfs,写完之后,开开森森的去交代码,却在第二个数据就TEL,然后优化半天,还是不行. 最终,我盯着1,2,5发呆半天,wc,然后直接贪心 #include<iostream> using namespace std; int main() { int T; cin>>T; while(T--) { int a,b; cin>>a>>b; int x = abs(a - b); ; x %= ; ans += x /…

先分解质因数,对于当前a[i],假设当前的质因数为x,这个位置要满足能被k整除,有两个可能,要么是它向后一个转移x%k个,要么是后一个向它转移k-x%k个. 对于每一个a[i]满足后,因为只会对下一个位置产生影响,所以下一个位置a[i+1]算上a[i]产生的影响,之后又是一个新的子问题(禁止套娃). 对于每一个质因数x,对所有所有位置i求min(a[i],x-a[i])的和,取所有质因子i的求和的最小值作为答案. #define HAVE_STRUCT_TIMESPEC #include<bit…

#define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; vector<char>ans; ][]; int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); ';++i) ans.emplace_back(i); for(int i='A';i<='Z';++i) ans.emplace_back(i…

A. Vitya in the Countryside time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Every summer Vitya comes to visit his grandmother in the countryside. This summer, he got a huge wart. Every gran…

A. Hongcow Learns the Cyclic Shift time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Hongcow is learning to spell! One day, his teacher gives him a word that he needs to learn to spell. Bein…

A. One-dimensional Japanese Crossword time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a tabl…

A. Crazy Computer time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output ZS the Coder is coding on a crazy computer. If you don't type in a word for a c consecutive seconds, everything you typed…

题目链接: http://codeforces.com/contest/670/problem/E 题解: 用STL的list和stack模拟的,没想到跑的还挺快. 代码: #include<iostream> #include<cstdio> #include<cstring> #include<map> #include<list> #include<stack> using namespace std; + ; const in…

题目链接:http://codeforces.com/contest/832/problem/B B. Petya and Exam time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output It's hard times now. Today Petya needs to score 100 points on Informatics…

A. Vasya and Football 题目连接: http://codeforces.com/contest/493/problem/A Description Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player wh…

A. Inna and Choose Options time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There always is something to choose from! And now, instead of "Noughts and Crosses", Inna choose a very unusu…

http://codeforces.com/contest/816/problem/A 题意: 给出一个时间,问最少过多少时间后是回文串. 思路: 模拟,先把小时的逆串计算出来: ① 如果逆串=分钟,那么此时已经是回文串了. ② 如果逆串>分钟,那么只需要逆串-分钟即可.(注意此时逆串>=60的情况) ③ 如果逆串<分钟,此时在这个小时内要构成回文串已经是不可能的了,那么就加上60-minute分钟,进入一个新的小时,然后重复上述步骤. #include<iostream>…

A. Combination Lock time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there…

传送门:http://codeforces.com/contest/1105/problem/B B. Zuhair and Strings time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Given a string ss of length nn and integer kk (1≤k≤n1≤k≤n). The string…

D. Pair of Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l…

题目 大致题意 n表示要进行n次操作,接着给出三个字符串,表示三个人初始拥有的串.每次操作要替换字符串中的字母,询问最后在游戏中曾出现过的相同的子串谁最多. 思路 (1)  讨论最多的子串,肯定是全部转换成单个的字母是最优的,这样就把子串转换成了讨论出现过最多的字母的问题.接下来只需要模拟,先将初始串中有的最多的字母数统计出来,然后考虑一下剩下的字符数和n的关系就可以了.需要注意的是,初始的字符串是不计入统计的,也就是说,至少在经过一次操作后我们才对它们的最大长度进行比较. (2)显然只需关注字…

http://codeforces.com/contest/723/problem/C 题目是给出一个序列 a[i]表示第i个歌曲是第a[i]个人演唱,现在选出前m个人,记b[j]表示第j个人演唱歌曲的数量, 现在你可以改变a[]这个数组,使得前m个人的演唱歌曲的数量的最小值最大. 很明显对于前m个人,每个人唱了多少首是很容易统计出来的,只需要对<=m的人进行统计即可,不用统计>m的,这样的话数组只需开到2000即可. 对于后面的人,可以改变,优先改变前m个人中演唱歌曲最小的那个,这个可以用优…