威佐夫博弈 POJ 1067

如果xy中较小的数等于xy的差值乘黄金分割比的话说先手必败
否则先手必胜

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代码如下：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int a,b;
signed main(void){
    double k=(sqrt(5.0)+1.0)/2.0;//黄金分割比 
    while(scanf("%d%d",&a,&b)!=EOF){
        if(min(a,b)==(int)(abs(a-b)*k))
            cout<<0<<endl;
        else
            cout<<1<<endl;
    }
    return 0;
}

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by >o< neighthorn