HDOJ1527博弈论之Wythoff游戏

Posted 2020-11-28 randy-lo

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1527

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef unsigned int ui;
typedef long long ll;
typedef unsigned long long ull;
#define pf printf
#define mem(a,b) memset(a,b,sizeof(a))
#define prime1 1e9+7
#define prime2 1e9+9
#define pi 3.14159265
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define scand(x) scanf("%llf",&x) 
#define f(i,a,b) for(int i=a;i<=b;i++)
#define scan(a) scanf("%d",&a)
#define mp(a,b) make_pair((a),(b))
#define P pair<int,int>
#define dbg(args) cout<<#args<<":"<<args<<endl;
#define inf 0x7ffffff
inline int read(){
    int ans=0,w=1;
    char ch=getchar();
    while(!isdigit(ch)){if(ch==‘-‘)w=-1;ch=getchar();}
    while(isdigit(ch))ans=(ans<<3)+(ans<<1)+ch-‘0‘,ch=getchar();
    return ans*w;
}
int n,m,t;
const int maxn=1e5+10;
const ll mod=10000;
int main()
{
//    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    double gold=(1+sqrt(5))/2;
    while(cin>>n>>m){
        int a=min(n,m);
        int b=max(n,m);
        double k=(double)(b-a);
        int test=(int)(k*gold);//每个奇异局势的较小值一定是等于差值*黄金分割比向下取整 
        if(test==a)cout<<0<<endl;//判断是否是奇异局势，奇异局势之下先手必败 
        else cout<<1<<endl;
    } 
}