UVA - 11542 Square (异或方程组)

Given n integers you cangenerate 2ⁿ-1 non-empty subsets from them. Determine for howmany of these subsets the product of all the integers in that is a perfectsquare. For example for the set {4,6,10,15} there are3 such subsets. {4}, {6,10,15} and {4,6,10,15}. Aperfect square is an integer whose square root is an integer. For example 1, 4,9, 16,…. .

 

Input

 

Input contains multiple testcases. First line of the input contains T(1≤T≤30)the number of test cases. Each test case consists of 2 lines. First line containsn(1≤n≤100) and second linecontainsn space separated integers. All these integers are between 1 and 10^15.  None of these integers is divisible by aprime greater than 500.

 

Output

 

For each test caseoutput is a single line containing one integer denoting the number of non-emptysubsets whose integer product is a perfect square. The input will be such thatthe result will always fit into signed 64 bit integer.

 

SampleInput                              Output for Sample Input

4

3

2 3 5

3

6 10 15

4

4 6 10 15

3

2 2 2

0

1

3

3

 

Problemsetter: Abdullah al Mahmud

SpecialThanks to: Manzurur RahmanKhan

题意：给出n个正整数，从中选出1个或者多个，使得选出来的整数乘积是完全平方数，一共有多少种选法。

思路：用01向量表示一个数，再用n个01向量来表示我们的选择，因为完全平方数要求素因子的次数一定要是偶数的，所以我们可以统计的将奇数当作1，偶数当作0，那么就是一组可以变换成oxr的方程组，最后的结果有自由变量f个，答案是2^f-1,f求解就是求n-方程组的秩

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <vector>typedef long long ll;using namespace std;const int maxn = 510;typedef int Matrix[maxn][maxn];int prime[maxn], vis[maxn];Matrix A;int gen_primes(int m) {memset(vis, 0, sizeof(vis));int cnt = 0;for (int i = 2; i < m; i++) {if (!vis[i]) {prime[cnt++] = i;for (int j = i * i; j < m; j += i)vis[j] = 1;}}return cnt;}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;}int main() {int m = gen_primes(505);int t;scanf("%d", &t);while (t--) {int n, maxp = 0;;ll x;scanf("%d", &n);memset(A, 0, sizeof(A));for (int i = 0; i < n; i++) {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("%lld\n", (1ll << (n-r)) - 1);}return 0;}