HDU 1525 - Euclid's Game-优快云博客

本文介绍了一个基于数学原理的算法游戏——欧几里得游戏。游戏中两名玩家交替操作，通过减法策略来减少两个自然数之一的值，直至一方能够使其中一个数变为0即获胜。文章提供了游戏规则说明、示例过程及代码实现。

Euclid's Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1166 Accepted Submission(s): 537

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

============================

思路：http://www.cnblogs.com/kuangbin/archive/2013/05/04/3059871.html

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int main()
{
    int a,b;
    while(~scanf("%d%d",&a,&b))
    {
        if(a==0&&b==0)
            break;
        bool flag=true;
        while(1)
        {
            if(a<b)
            {
                a^=b;b^=a;a^=b;
            }
            if(a/b>=2||a==b)
                break;
            a-=b;
            flag=!flag;
        }
        if(flag)
        {
            cout<<"Stan wins"<<endl;
        }
        else
        {
            cout<<"Ollie wins"<<endl;
        }
    }
    return 0;
}