POJ 3553 Light Switching Game 博弈论 nim积 sg函数

Posted 2020-10-22 137shoebills

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

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<map>
 7 #include<ctime>
 8 using namespace std;
 9 long long n;
10 long long sg[21][21]={};
11 long long f(long long,long long);
12 long long g(long long x,long long y){
13     if(sg[x][y]!=-1)return sg[x][y];
14     if(!x)return sg[x][y]=1<<y;
15     if(!y)return sg[x][y]=1<<x;
16     long long ans=1,k=1,t;
17     long long x1=x,y1=y;
18     while(x||y){
19         t=1<<k;
20         if((x^y)&1){
21             ans*=t;
22         }
23         x>>=1;y>>=1;k<<=1;
24     }
25     k=1;x=x1;y=y1;
26     while(x||y){
27         t=1<<k;
28         if((x&y)&1){
29             ans=f(ans,t/2*3);
30         }
31         x>>=1;y>>=1;k<<=1;
32     }return sg[x1][y1]=ans;
33 }
34 long long f(long long x,long long y){
35     if(!x||!y)return 0;
36     if(x==1)return y;
37     if(y==1)return x;
38     long long ans=0;
39     for(long long i=x,a=0;i;i>>=1,a++){
40         if(!(i&1))continue;
41         for(long long j=y,b=0;j;j>>=1,b++){
42             if(!(j&1))continue;
43             ans^=g(a,b);
44         }
45     }return ans;
46 }
47 int main(){
48     memset(sg,-1,sizeof(sg));
49     while(~scanf("%lld",&n)){
50         long long ans=0,x,y,z;
51         for(long long i=1;i<=n;i++){
52             scanf("%lld%lld%lld",&x,&y,&z);
53             ans^=f(z,f(x,y));
54         }
55         if(ans)printf("No\n");
56         else printf("Yes\n");
57     }
58     return 0;
59 }

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