[简单博弈] hdu 1525 Euclid's Game

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1525

Euclid's Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1832    Accepted Submission(s): 808

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
   
   
   
   

 

Source

University of Waterloo Local Contest 2002.09.28

 

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题目意思：

两个人，两堆石头，玩游戏。每次从石头数多的那堆中拿走石头数少的那堆石头的个数的倍数个，保证拿了过后不能为负数。谁恰好使其中一堆石头数为0，谁使其中某一堆石头为0，谁赢。两人都足够聪明。

解题思路：

借助欧几里得算法的博弈。

显然如果a>b 且a>2*b  先拿者必赢 因为他可以决定是自己还是对手遇到状态（a%b,b）如果状态（a%b,b）为必输状态，他就让对手面对（拿走a-a%b个）。如果为必赢状态，他就自己面对(拿走-a%b-b个).

如果b<a<=2*b 只能到达（a%b,b)的状态。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

int gcd(ll a,ll b)
{
    if(a<b) //求出谁大
        swap(a,b);
    if(b==0) //终止 必输状态
        return 0;
    if(a>2*b) //必胜
        return 1;

    return gcd(b,a%b)^1; //和拿了之后的输赢情况相反
}

int main()
{
   //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   ll a,b;

   while(scanf("%I64d%I64d",&a,&b))
   {
       if(!a&&!b)
          break;
       int ans=gcd(a,b);

       if(ans)
            printf("Stan wins\n");
       else
            printf("Ollie wins\n");
   }
   return 0;
}