OpenJudge noi 1350Euclid's Game

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

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

这题目大致题意就是说，给你两个数，两个人依次对较大的那个数进行修改。修改的方法是把较大的数减去较小的数的任意正整数倍数，直到谁修改到0，谁就赢了。这个东西可以用fiber来证明。我们知道，对于fiber它的比例随着项数的增大而逐渐接近（√5-1）/2，而且相邻两项会依次小于，大于这个数。我们可以判断这个数是大于还是小于这个数，从而可以判断先手还是后手才赢。如果m==n，则先手会必胜，这个显然

因此，代码如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
using namespace std;
const double p=(sqrt(5)-1)/2;
int main()
{
    double m,n;
    while(true)
    {
        cin>>m>>n;
        if(m==0&&n==0)
        break;
        if(m==n)
            printf("Stan wins\n");
        else
        {
            if(n>m) swap(n,m);
            double c=n/m;
            if(c>p)
            printf("Ollie wins\n");
            else
            printf("Stan wins\n");
        }       
    }
    return 0；
}

还有一种方法是辗转相除法。