Light OJ 1288 subsets Forming Perfect squares Gaussian elimination to calculate the rank of matrix

Source: Internet
Author: User
Tags cas

Source: Light OJ 1288 subsets Forming Perfect Squares

Test instructions: Give you the number of N to select some numbers their product is the total squared number of how many schemes

Idea: Each number decomposition factor can be selected or not to choose 0 1 means that the number of elements selected by the M prime factor must be an even

Each row of matrices that comprise a M-row and N-column represents the number of free elements for each factor's coefficients

#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];
	return a^p mod n fast power 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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