(HDUStep 1.2.2)hide handkerchief(用辗转相除法来求最大公约数)-CFANZ编程社区

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
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题目分析:这中体实际上判断一下输入的两个数n,m是否互质就行了。如果在使用辗转相除法求得的最大公约数是1，则证明是互质，否则就不是互质

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;
}