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hide handkerchief |
| Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 5050 Accepted Submission(s): 1676 |
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 people played the game ,who sit on the ground forming a circle ,everyone owns a box behind them .Also there is a beautiful handkerchief hid in a box which is one of the boxes . Then Haha(a friend of mine) is called to find the handkerchief. But he has a strange habit. Each time he will search the next box which is separated by M-1 boxes from the current box. For example, there are three boxes named A,B,C, and now Haha is at place of A. now he decide the M if equal to 2, so he will search A first, then he will search the C box, for C is separated by 2-1 = 1 box B from the current box A . Then he will search the box B ,then he will search the box A. So after three times he establishes that he can find the beautiful handkerchief. Now I will give you N and M, can you tell me that Haha is able to find the handkerchief or not. If he can, you should tell me "YES", else tell me "POOR Haha". |
Input There will be several test cases; each case input contains two integers N and M, which satisfy the relationship: 1<=M<=100000000 and 3<=N<=100000000. When N=-1 and M=-1 means the end of input case, and you should not process the data. |
Output For each input case, you should only the result that Haha can find the handkerchief or not. |
Sample Input 3 2
-1 -1 |
Sample Output YES |
Source HDU 2007-6 Programming Contest |
Recommend xhd |
关于辗转相除法求最大公约数的流程：
典型例题： 一.辗转相除法 例1 。求两个正数8251和6105的最大公因数。 （分析：辗转相除→余数为零→得到结果） 解：8251＝6105×1＋2146 显然8251与6105的最大公因数也必是2146的因数，同样6105与2146的公因数也必是8251的因数，所以8251与6105的最大公因数也是6105与2146的最大公因数。 6105＝2146×2＋1813 2146＝1813×1＋333 1813＝333×5＋148 333＝148×2＋37 148＝37×4＋0 则37为8251与6105的最大公因数。 以上我们求最大公因数的方法就是辗转相除法。也叫欧几里德算法，它是由欧几里德在公元前300年左右首先提出的。 1． 为什么用这个算法能得到两个数的最大公因数？ 利用辗转相除法求最大公因数的步骤如下： 第一步：用较大的数m除以较小的数n得到一个商q0和一个余数r0； 第二步：若r0＝0，则n为m，n的最大公因数；若r0≠0，则用除数n除以余数r0得到一个商q1和一个余数r1； 第三步：若r1＝0，则r1为m，n的最大公因数；若r1≠0，则用除数r0除以余数r1得到一个商q2和一个余数r2； …… 依次计算直至rn＝0，此时所得到的rn－1即为所求的最大公因数。
代码如下：
/*
* b.cpp
*
* Created on: 2015年1月27日
* Author: Administrator
*/
#include <iostream>
#include <cstdio>
using namespace std;
/**
* 求最大公约数
*/
int gcd(int a,int b){
int temp;
while(b != 0){
temp = b;
b = a%b;
a = temp;
}
return a;
}
int main(){
int n,m;
while(scanf("%d%d",&n,&m)!=EOF,n!=-1){
if(gcd(n,m) == 1){
printf("YES\n");
}else{
printf("POOR Haha\n");
}
}
return 0;
}
(HDUStep 1.2.2)hide handkerchief(用辗转相除法来求最大公约数)
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原文地址：http://blog.csdn.net/hjd_love_zzt/article/details/43194213
