HDU 2104 hide handkerchief

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HDU 2104 hide handkerchief

题解:由题目可以知道,如果n和m的最大公约数不为1,那么总有箱子是无法遍历的,所以求一遍GCD就可以判断了. 注意点:一定要记住判断是==,在做题时又忘了. #include <cstdio> int gcd(int a,int b) { ) return(a); else return(gcd(b,(a%b))); } int main() { int n,m; ||m!=-) { ) puts("YES"); else puts("POOR Haha"…

【HDOJ】2104 hide handkerchief

Problem Description The Children’s Day has passed for some days .Has you remembered something happened at your childhood? I remembered I often played a game called hide handkerchief with my friends.Now I introduce the game to you. Suppose there are N…

hdu 2104

#include <iostream> using namespace std; int gcd(int a,int b) { return (b?gcd(b,a%b):a); } int main() { int m,n; while(cin>>m>>n&&m+n!=-2) { if(gcd(m,n)==1) cout<<"YES\n"; else cout<<"POOR Haha\n";…

hide handkerchief

Problem Description The Children’s Day has passed for some days .Has you remembered something happened at your childhood? I remembered I often played a game called hide handkerchief with my friends.Now I introduce the game to you. Suppose there are N…

hdu 2104(判断互素)

hide handkerchief Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3970 Accepted Submission(s): 1884 Problem Description The Children’s Day has passed for some days .Has you remembered someth…

数学--数论--HDU 2104 丢手绢（离散数学 mod N+ 剩余类生成元）+（最大公约数）

The Children's Day has passed for some days .Has you remembered something happened at your childhood? I remembered I often played a game called hide handkerchief with my friends. Now I introduce the game to you. Suppose there are N people played the…

HDU 1941 Hide and Seek（离散化+树状数组）

题目链接:http://61.187.179.132/JudgeOnline/problem.php?id=1941 题意:给出平面上n个点,找出一点p,使得距离p最近和最远的点的距离之差最小.输出这个差值. 思路: 计算出距离每个点最近的点的距离和最远点的的距离.怎么计算呢?首先我们计算每个点到其左下角的那些点的最近最远距离.只要按照x升序然后y升序,之后将y离散化.一个一个插入到树状数组中.这样的话(x2+y2)-(x1+y1)的差值就是它们的距离.对于其右下角.左上角.右上角乘以(-1,1…