题目链接: C1. Brain Network (easy) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output One particularly well-known fact about zombies is that they move and think terribly slowly. While we still do…
题意:给定 n 条边,判断是不是树. 析:水题,判断是不是树,首先是有没有环,这个可以用并查集来判断,然后就是边数等于顶点数减1. 代码如下: #include <bits/stdc++.h> using namespace std; const int maxn =1000 + 5; int p[maxn]; int Find(int x){ return x == p[x] ? x : p[x] = Find(p[x]); } int main(){ int n, m, x, y; cin…
题意:给一张照片的像素,让你来确定是黑白的还是彩色的. 析:很简单么,如果有一种颜色不是黑白灰,那么就一定是彩色的. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #i…
G - Brain Network (easy) Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u CodeForces 690C1 Description One particularly well-known fact about zombies is that they move and think terribly slowly. While we still don't know…
Brain Network (easy) One particularly well-known fact about zombies is that they move and think terribly slowly. While we still don't know why their movements are so sluggish, the problem of laggy thinking has been recently resolved. It turns out tha…
A. Brain's Photos 题目连接: http://www.codeforces.com/contest/707/problem/A Description Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but t…
这题一眼看就是水题,map随便计 然后我之所以发这个题解,是因为我用了log2()这个函数判断在哪一层 我只能说我真是太傻逼了,这个函数以前听人说有精度问题,还慢,为了图快用的,没想到被坑惨了,以后尽量不用 #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> #…
题目链接: B. Modulo Sum time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given a sequence of numbers a1, a2, ..., an, and a number m. Check if it is possible to choose a non-empty subse…
A.Beru-taxi 水题:有一个人站在(sx,sy)的位置,有n辆出租车,正向这个人匀速赶来,每个出租车的位置是(xi, yi) 速度是 Vi;求人最少需要等的时间: 单间循环即可: #include<iostream> #include<algorithm> #include<string.h> #include<stdio.h> #include<math.h> #include<vector> using namespace…
题意:给定 n 和 k,然后是 n 个数,表示1-k的一个值,问你修改最少的数,使得所有的1-k的数目都等于n/k. 析:水题,只要用每个数减去n/k,然后取模,加起来除以2,就ok了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cma…