hdu1525-------Euclid's Game 找规律

Euclid's Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 893 Accepted Submission(s): 419

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

Recommend
LL
题意：给你两个数，每次可以拿大的那个数减去小的那个数的正整数倍，只要减去后得到的数是正的或者0就行，谁先得到其中一个数是0，谁就胜出。
我们会发现，假设a>b,如果a/b>=2,那么后面就会出现a/b种路线，当a/b=1的时候只有一种路线，所以谁到了a/b>=2这个局面就有必胜策略，此外，当a%b==0的时候，就直接跳出来了，也是必胜点。
[cpp]
#include<iostream>
#include<cstdlib>
#include<stdio.h>
using namespace std;
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)&&(n&&m))
    {
       int count=1;
       while(1)
       {
           if(n<m) swap(n,m);
           if(n==0||m==0||(n/m>=2)||(n%m==0)) break;
           n%=m;
           count++;
       }
       if(count&1) puts("Stan wins");
        else puts("Ollie wins");
    }
}

补充：软件开发 , C++ ,