UVa 11542 (高斯消元 异或方程组) Square

书上分析的太清楚，我都懒得写题解了。=_=||

 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 
 const int maxp = ;
 const int maxn = ;
 bool vis[maxn + ];
 int prime[maxp], pcnt = ;
 
 void Init()
 {
     int m = sqrt(maxn + 0.5);
     for(int i = ; i <= m; i++) if(!vis[i])
         for(int j = i*i; j <= maxn; j += i) vis[j] = true;
     for(int i = ; i <= maxn; i++) if(!vis[i]) prime[pcnt++] = i;
 }
 
 typedef int Matrix[maxn][maxn];
 
 Matrix A;
 
 int rank(Matrix A, int m, int n)
 {//求系数矩阵A的秩，m个方程，n个未知数
     int i = , j = ;
     while(i < m && j < n)
     {
         int r = i, k;
         for(k = r; k < m; k++) if(A[k][j]) { r = k; break; }
         if(k < m)
         {
             if(r != i) for(int k = ; k < n; k++) swap(A[r][k], A[i][k]);
             for(int k = i+; k < m; k++) if(A[k][j])
                 for(int l = j; l < n; l++) A[k][l] ^= A[i][l];
             i++;
         }
         j++;
     }
     return i;
 }
 
 int main()
 {
     //freopen("in.txt", "r", stdin);
 
     Init();
     int T;
     scanf("%d", &T);
     while(T--)
     {
         memset(A, , sizeof(A));
         int n, M = ;
         scanf("%d", &n);
         for(int i = ; i < n; i++)
         {
             long long x;
             scanf("%lld", &x);
             for(int j = ; j < pcnt; j++) while(x % prime[j] == )
             {
                 M = max(M, j);
                 x /= prime[j];
                 A[j][i] ^= ;
             }
         }
         int r = rank(A, M+, n);//共用到前M+1个素数
         printf("%lld\n", (1LL << (n-r)) - );
     }
 
     return ;
 }

代码君

最后lrj老师提到了还可以用状压加速消元，因为500以内的素数不超过100个，所以我用了两个64位的long long来表示一个方程。第一份代码16ms，状压以后12ms，快了四分之一。

 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 
 const int maxp = ;
 const int maxn = ;
 bool vis[maxn + ];
 int prime[maxp], pcnt = ;
 
 void Init()
 {
     int m = sqrt(maxn + 0.5);
     for(int i = ; i <= m; i++) if(!vis[i])
         for(int j = i*i; j <= maxn; j += i) vis[j] = true;
     for(int i = ; i <= maxn; i++) if(!vis[i]) prime[pcnt++] = i;
 }
 
 typedef long long Matrix[maxn][];
 
 Matrix A;
 
 int rank(Matrix A, int m, int n)
 {//求系数矩阵A的秩，m个方程，n个未知数
     int i = , j = , len = n / ;
     while(i < m && j < n)
     {
         int r = i, k;
         for(k = r; k < m; k++) if(A[k][j/] & (1LL<<(j%))) { r = k; break; }
         if(k < m)
         {
             if(r != i) for(int k = ; k <= len; k++) swap(A[r][k], A[i][k]);
             for(int k = i+; k < m; k++) if(A[k][j/] & (1LL<<(j%)))
                 for(int l = ; l <= len; l++) A[k][l] ^= A[i][l];
             i++;
         }
         j++;
     }
     return i;
 }
 
 int main()
 {
     //freopen("in.txt", "r", stdin);
 
     Init();
     int T;
     scanf("%d", &T);
     while(T--)
     {
         memset(A, , sizeof(A));
         int n, M = ;
         scanf("%d", &n);
         for(int i = ; i < n; i++)
         {
             long long x;
             scanf("%lld", &x);
             for(int j = ; j < pcnt; j++) while(x % prime[j] == )
             {
                 M = max(M, j);
                 x /= prime[j];
                 A[j][i/] ^= (1LL << (i%) );
             }
         }
         int r = rank(A, M+, n);//共用到前M+1个素数
         printf("%lld\n", (1LL << (n-r)) - );
     }
 
     return ;
 }

代码君