[poj1830]开关问题（高斯消元）

题意：求高斯消元中自由元的个数，输出1<<ans;

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 const int N=;
 int a[N][N],ans[N];
 int T,n,x,nn;
 int gauss(int nn){
     int i,j,k,l;
     for(i=,j=;i<=nn&&j<=nn;j++){
         for(k=i;k<=nn;k++)if(a[k][j])break;
         if(a[k][j]){
             for(l=;l<=nn+;l++)swap(a[i][l],a[k][l]);
             for(l=;l<=nn;l++){
                 if(l!=i&&a[l][j])for(k=;k<=nn+;k++)a[l][k]^=a[i][k];
             }
             i++;
         }
     }
     for(j=i;j<=nn;j++)if(a[j][n+])return -;
     return <<(n-i+);
 }

 int main(){
     scanf("%d",&T);
     while(T--){
         memset(a,,sizeof a);
         scanf("%d",&n);
         for(int i=;i<=n;i++)scanf("%d",&a[i][n+]);
         for(int i=;i<=n;i++)scanf("%d",&x),a[i][n+]^=x;
         int tx,ty;
         while(scanf("%d%d",&tx,&ty)&&(tx||ty))a[ty][tx]=;//注意
         for(int i=;i<=n;i++)a[i][i]=;
         int ans=gauss(n);
         if(ans==-)printf("Oh,it's impossible~!!\n");
         else printf("%d\n",ans);
     }
     return ;
 }