poj 2348 Euclid's Game-----博弈

最新推荐文章于 2020-10-29 20:26:38 发布

iteye_4476

最新推荐文章于 2020-10-29 20:26:38 发布

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文章标签： c/c++

本文详细解读了欧几里得游戏的策略，通过分析游戏规则和最优解法，揭示了如何在有限的回合内赢得游戏。文章深入探讨了数学逻辑在游戏中的应用，为读者提供了理解和解决类似博弈问题的思路。

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Euclid's Game

Time Limit:1000MS		Memory Limit:65536K
Total Submissions:6030		Accepted:2441

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):

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

题目 没难度， 但是被坑了，int型不够。  所以要 long long 。

题目大意：  每次可以 把两个数中 较大数 减去 较小数的 倍数   ，  两人轮流操作， 先到 0 的赢。

//Memory: 164 KB		Time: 0 MS
//Language: C++		Result: Accepted
#include<stdio.h>
#define LL long long
int main()
{
//	freopen("in.txt","r",stdin);
	LL m,n,t,ok;
	while(~scanf("%lld%lld",&m,&n) && (n||m))
	{
		ok=1;
		if(n>m)t=m,m=n,n=t;
		while(1)
		{
			if(m>=2*n) break;
			m-=n;
			if(n>m)t=m,m=n,n=t;
			if(!n) break;
			ok=(!ok);
		}
		if(ok) printf("Stan wins\n");
		else printf("Ollie wins\n");	
	}
	return 0;
}