POJ 3553 Light Switching Game 博弈论 nim积 sg函数

http://poj.org/problem?id=3533

变成三维的nim积。。前面hdu那个算二维nim积的题的函数都不用改，多nim积一次就过了。。。longlong似乎不必要但是还是加上了

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<map>
#include<ctime>
using namespace std;
long long n;
long long sg[21][21]={};
long long f(long long,long long);
long long g(long long x,long long y){
    if(sg[x][y]!=-1)return sg[x][y];
    if(!x)return sg[x][y]=1<<y;
    if(!y)return sg[x][y]=1<<x;
    long long ans=1,k=1,t;
    long long x1=x,y1=y;
    while(x||y){
        t=1<<k;
        if((x^y)&1){
            ans*=t;
        }
        x>>=1;y>>=1;k<<=1;
    }
    k=1;x=x1;y=y1;
    while(x||y){
        t=1<<k;
        if((x&y)&1){
            ans=f(ans,t/2*3);
        }
        x>>=1;y>>=1;k<<=1;
    }return sg[x1][y1]=ans;
}
long long f(long long x,long long y){
    if(!x||!y)return 0;
    if(x==1)return y;
    if(y==1)return x;
    long long ans=0;
    for(long long i=x,a=0;i;i>>=1,a++){
        if(!(i&1))continue;
        for(long long j=y,b=0;j;j>>=1,b++){
            if(!(j&1))continue;
            ans^=g(a,b);
        }
    }return ans;
}
int main(){
    memset(sg,-1,sizeof(sg));
    while(~scanf("%lld",&n)){
        long long ans=0,x,y,z;
        for(long long i=1;i<=n;i++){
            scanf("%lld%lld%lld",&x,&y,&z);
            ans^=f(z,f(x,y));
        }
        if(ans)printf("No
");
        else printf("Yes
");
    }
    return 0;
}