hdu 1525 Euclid's Game【 博弈论】

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.

InputThe input consists of a number of lines. Each line contains two positive integers giving the starting two numbers of the game. Stan always starts.
OutputFor 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

如果a % b == 0 先手必胜

如果a >= 2b 那么如果a % b,b是必败态，先手可以把数字改为a % b, b 如果是必胜态 就可以把数字变为a % b + b, b而后手只能操作变成a % b, b 先手必胜

如果a < 2b 就一直减 直到出现必胜态

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<limits>
#include<stack>
#include<queue>
#include<cmath>
#define inf 1000005
using namespace std;
int a, b;
int main()
{
    while(scanf("%d%d",&a,&b) != EOF && (a || b)){
        if(a < b)
            swap(a,b);
        if(!(a % b) || a >= 2 * b){
            printf("Stan wins\n");
        }
        else{
            int k = 0;
            while(a < 2 * b && a % b){
                a -= b;
                if(a < b){
                    swap(a,b);
                }
                ++k;
            }
            if(k % 2){
                printf("Ollie wins\n");
            }
            else{
                printf("Stan wins\n");
            }
        }
    }
    return 0;
}