題意：求高斯消元中自由元的個數，輸出1<<ans;

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<algorithm>
 7 using namespace std;
 8 const int N=35;
 9 int a[N][N],ans[N];
10 int T,n,x,nn;
11 int gauss(int
 
 nn){
12     int i,j,k,l;
13     for(i=1,j=1;i<=nn&&j<=nn;j++){
14         for(k=i;k<=nn;k++)if(a[k][j])break;
15         if(a[k][j]){
16             for(l=1;l<=nn+1;l++)swap(a[i][l],a[k][l]);
17             for(l=1;l<=nn;l++){
18                 if(l!=i&&a[l][j])for
 
(k=1;k<=nn+1;k++)a[l][k]^=a[i][k];
19             }
20             i++;
21         }
22     }
23     for(j=i;j<=nn;j++)if(a[j][n+1])return -1;
24     return 1<<(n-i+1); 
25 }
26 
27 int main(){
28     scanf("%d",&T);
29     while(T--){
30         memset(a,0,sizeof a);
31         scanf("
 
%d",&n);
32         for(int i=1;i<=n;i++)scanf("%d",&a[i][n+1]);
33         for(int i=1;i<=n;i++)scanf("%d",&x),a[i][n+1]^=x;
34         int tx,ty;
35         while(scanf("%d%d",&tx,&ty)&&(tx||ty))a[ty][tx]=1;//註意 
36         for(int i=1;i<=n;i++)a[i][i]=1;
37         int ans=gauss(n);
38         if(ans==-1)printf("Oh,it‘s impossible~!!\n");
39         else printf("%d\n",ans); 
40     }
41     return 0;
42 }

[poj1830]開關問題（高斯消元）