Euclid's Game - POJ 2348 博弈论_euclid'game-优快云博客

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/46128053

本文介绍了一个基于数学原理的游戏——欧几里得游戏，玩家通过从较大数中减去较小数的倍数来进行博弈。文章提供了算法思路及AC代码实现，并分析了胜负条件。

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Euclid's Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8066		Accepted: 3277

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

题意：设两堆石子数量为a和b，a>=b，每次在a堆里取b的倍数个石子，出现0则不能取并且输掉游戏。问先手胜负。

思路：如果a/b>=2则先手必胜，否则为b,a%b的情况的取反。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int solve(ll a,ll b)
{
    if(a<b)
      swap(a,b);
    if(b==0)
      return 0;
    if(a/b>=2)
      return 1;
    else
      return solve(a%b,b)^1;
}
int main()
{
     int i,j,k;
     ll a,b;
     while(~scanf("%I64d%I64d",&a,&b) && a+b>0)
     {
         if(solve(a,b))
           printf("Stan wins\n");
         else
           printf("Ollie wins\n");
     }
}