Light OJ 1288 Subsets Forming Perfect Squares 高斯消元求矩阵的秩

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题目来源：Light OJ 1288 Subsets Forming Perfect Squares

题意：给你n个数选出一些数他们的乘积是完全平方数求有多少种方案

思路：每个数分解因子每隔数可以选也可以不选 0 1表示然后设有m种素数因子选出的数组成的各个因子的数量必须是偶数

组成一个m行和n列的矩阵每一行代表每一种因子的系数解出自由元的数量

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn = 1010;const int mod = 1000000007;typedef int Matrix[maxn][maxn];typedef long long LL;int prime[maxn];bool vis[maxn];//返回a^p mod n 快速幂LL pow_mod(LL a, LL p, LL n){LL ans = 1;while(p){if(p&1){ans *= a;ans %= n;}a *= a;a %= n;p >>= 1;}return ans;}void sieve(int n){int m = sqrt(n+0.5);memset(vis, 0, sizeof(vis));vis[0] = vis[1] = 1;for(int i = 2; i <= m; i++)if(!vis[i])for(int j = i*i; j <= n; j += i)vis[j] = 1;}int get_primes(int n){sieve(n);int c = 0;for(int i = 2; i <= n; i++)if(!vis[i])prime[c++] = i;return c;}int rank(Matrix A, int m, int n){int i = 0, j = 0, k, r, u;while(i < m && j < n){r = i;for(k = i; k < m; k++)if(A[k][j]){r = k;break;}if(A[r][j]){if(r != i)for(k = 0; k <= n; k++)swap(A[r][k], A[i][k]);for(u = i+1; u < m; u++)if(A[u][j])for(k = i; k <= n; k++)A[u][k] ^= A[i][k];i++;}j++;}return i;}Matrix A;int main(){int cas = 1;int m = get_primes(500);int T;scanf("%d", &T);while(T--){int n, maxp = 0;scanf("%d", &n);memset(A, 0, sizeof(A));for(int i = 0; i < n; i++){long long x;scanf("%lld", &x);for(int j = 0; j < m; j++){while(x % prime[j] == 0){maxp = max(maxp, j);x /= prime[j];A[j][i] ^= 1;}}}int r = rank(A, maxp+1, n);printf("Case %d: %lld\n", cas++, pow_mod(2, n-r, mod)-1);}return 0;}

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