Codeforces Round #270 A. Design Tutorial: Learn from Math time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output One way to create a task is to learn from math. You can generate some random math s…

Codeforces Round #270 1003 C. Design Tutorial: Make It Nondeterministic time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A way to make a new task is to make it nondeterministic or probabili…

Codeforces Round #270 1002 B. Design Tutorial: Learn from Life time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output One way to create a task is to learn from life. You can choose some experience…

Codeforces Round #270 1001 A. Design Tutorial: Learn from Math time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output One way to create a task is to learn from math. You can generate some random m…

D. Design Tutorial: Inverse the Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There is an easy way to obtain a new task from an old one called "Inverse the problem": we give…

题目链接:http://www.codeforces.com/problemset/problem/472/A题意:给你一个数n,将n表示为两个合数(即非素数)的和.C++代码: #include <iostream> using namespace std; bool isprime(int x) { ; i * i <= x; i ++) ) return false; return true; } int main() { int n; cin >> n; ;i <…

题目链接:http://codeforces.com/contest/472/problem/A 题目: 题意:哥德巴赫猜想是:一个大于2的素数一定可以表示为两个素数的和.此题则是将其修改为:一个大于等于12的数一定能表示为两个合数的和. 思路:这个很容易,下面是三种方法的代码. 奇偶法:一个数要么是奇数要么是偶数,众所周知大于2的偶数都是合数(因为都能被2整除嘛),所以只要把该数分解为两个非2的偶数的和即可. #include<bits/stdc++.h> using namespace s…

题意:给出一个距离矩阵,问是不是一颗正确的带权树. 解法:先按找距离矩阵建一颗最小生成树,因为给出的距离都是最短的点间距离,然后再对每个点跑dfs得出应该的dis[][],再对比dis和原来的mp是否一致即可. 首先还要判断一些东西.具体看代码吧. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #in…

谈论最激烈的莫过于D题了! 看过的两种做法不得不ORZ,特别第二种,简直神一样!!!!! 1th:构造最小生成树. 我们提取所有的边出来按边排序,因为每次我们知道边的权值>0, 之后每次把边加入集合中,不断构造,类似  kruskal算法,构造出边后 再对每个点进行整张图的DFS求距离 复杂度O(N^2lgN):对所有边排序的复杂度. #include<bits/stdc++.h> #define N 2222 using namespace std; typedef long long…

A 题意:给出一个数n,求满足a+b=n,且a+b均为合数的a,b 方法一:可以直接枚举i,n-i,判断a,n-i是否为合数 #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<stack> #include<vector> #include<map> #include<algorithm> #defin…