HDOj题目1525 Euclid's Game（博弈）

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Euclid's Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1997 Accepted Submission(s): 892

Problem Description

Two players, Stan and Ollie, play, starting with two natural numbers. Stan, the first player, subtracts any positive multiple of the lesser of the two numbers from the greater of the two numbers, provided that the resulting number must be nonnegative. Then Ollie, the second player, does the same with the two resulting numbers, then Stan, etc., alternately, until one player is able to subtract a multiple of the lesser number from the greater to reach 0, and thereby wins. For example, the players may start with (25,7):

25 7
11 7
4 7
4 3
1 3
1 0

an Stan wins.

Input

The input consists of a number of lines. Each line contains two positive integers giving the starting two numbers of the game. Stan always starts.

Output

For each line of input, output one line saying either Stan wins or Ollie wins assuming that both of them play perfectly. The last line of input contains two zeroes and should not be processed.

Sample Input

Sample Output

  
   Stan wins
Ollie wins

Source

University of Waterloo Local Contest 2002.09.28

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转载请注明出处，谢谢 http://blog.youkuaiyun.com/ACM_cxlove?viewmode=contents by---cxlove

给出两个数，a和b，将大的数中，减去若干b的倍数，最终有一个数为0的话就胜了。

http://acm.hdu.edu.cn/showproblem.php?pid=1525

两个数a和b，总会出现的一个局面是b,a%b，这是必然的，如果a>=b&&a<2*b，那么只有一种情况，直接到b,a%b。否则有多种情况。

对于对于a/b==1这种局面，只可能到b,a%b，没有选择。而a/b>=2的话，先手可以选择由谁面对b,a%b这样的局势

显然选手足够聪明b,a%b谁必胜必败已经清楚，先手在a/b>=2的局面必胜

ac 代码

#include<stdio.h>
int main()
{
	int a,b;
	
	while(scanf("%d%d",&a,&b)!=EOF&&a,b)
	{
		int stan=1;
		int t;
		if(a<b)
		{
			t=a;
			a=b;
			b=t;
		}
		while(1)
		{
			if(b==0||a%b==0||a/b>=2)
				break;
			t=a;
			a=b;
			b=t-a;
			stan=!stan;
		}
	 if(stan)  
            printf("Stan wins\n");  
        else  
            printf("Ollie wins\n");  
	}
}