hdu 1525 Euclid's Game 博弈~~需要分析一下

最新推荐文章于 2025-06-01 14:00:32 发布

Lionel_D

最新推荐文章于 2025-06-01 14:00:32 发布

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分类专栏： ACM 博弈论文章标签： hdu1525 ACM Euclids Game 博弈

本文链接：https://blog.youkuaiyun.com/Lionel_D/article/details/44418939

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本文介绍了一个基于数学原理的两人游戏——欧几里得游戏，并提供了详细的算法实现过程。游戏由两名玩家轮流进行，从两个初始正整数开始，通过特定规则操作直至一方获胜。文章还附带了完整的C语言实现代码。

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Euclid's Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2223 Accepted Submission(s): 1001

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

34 12
15 24
0 0

Sample Output

Stan wins
Ollie wins

看一个大神博客~

http://www.cnblogs.com/kuangbin/p/3204627.html

代码：

#include <stdio.h>

void swap(int &a , int &b)
{
	int temp = a ;
	a = b ;
	b = temp ;
}

int main()
{
	int n ;
	int a,b;
	while(~scanf("%d%d",&a,&b) && (a+b) )
	{
		if(a<b)
		{
			swap(a,b) ;
		}
		if(a/b>=2)
		{
			puts("Stan wins") ;
		}
		else
		{
			int win = 0 ;
			while(b)
			{
				if(a%b==0 || a/b >= 2)
					break ;
				a = a-b ;
				swap(a,b) ;
				win ^= 1 ;
			}
			if(win == 0)
				puts("Stan wins") ;
			else
				puts("Ollie wins") ;
		}
	}
	return 0 ;
}

与君共勉