HDU1525 Euclid's Game (找规律博弈)

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本文介绍了一个基于数学原理的游戏——欧几里得游戏，分析了游戏规则及获胜策略。通过递归减法的方式，两位玩家交替进行操作，目标是在合法的操作下使其中一个数变为0。文章给出了具体的实现代码，并通过样例输入输出展示了如何确定哪位玩家能赢得比赛。

Euclid's Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3434 Accepted Submission(s): 1599

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

思路：题目让求的是用大的数减去小的数，能够使得其中的一个数变成0的人为胜者。首先来分析下，假设a>=b，如果a==b 或者a%b==0则先手为胜，当a>=2*b时，则可以判断（a%b，b）是必胜点或者必败点，如果是必败点则可以走到这点，如果是必胜点则可以走到（a%b+b,b），当a < 2*b,则可以通过 a-=b计算

#include <cstdio>
#include <algorithm>
using namespace std;

int a,b;

int main(){
	while(~scanf("%d%d",&a,&b)){
		if(a==0&&b==0)
			break;
		if(a < b)
			swap(a,b);
		int flag = 0;
		while(b){
			if(a%b == 0 || a / b >= 2)
				break;
			a -= b;
			swap(a,b);
			flag = !flag;
		}
		if(flag)
			printf("Ollie wins\n");
		else
			printf("Stan wins\n");
	}
	return 0;
}