Zhu and 772002

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 71 Accepted Submission(s): 24

Problem Description
Zhu and 772002 are both good at math. One day, Zhu wants to test the ability of 772002, so he asks 772002 to solve a math problem.

But 772002 has a appointment with his girl friend. So 772002 gives this problem to you.

There are n numbers a1,a2,…,an. The value of the prime factors of each number does not exceed 2000, you can choose at least one number and multiply them, then you can get a number b.

How many different ways of choices can make b is a perfect square number. The answer maybe too large, so you should output the answer modulo by 1000000007.

Input
First line is a positive integer T , represents there are T test cases.

For each test case:

First line includes a number n(1≤n≤300)，next line there are n numbers a1,a2,…,an,(1≤ai≤1018).

Output
For the i-th test case , first output Case #i: in a single line.

Then output the answer of i-th test case modulo by 1000000007.

Sample Input
2
3
3 3 4
3
2 2 2

Sample Output
Case #1:
3
Case #2:
3

Author
UESTC

Source
2016中國大學生程式設計競賽 - 網路選拔賽

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原題醉了。。CCPC網路賽居然出原題。。UVA11542用亦或高斯消元一下

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
typedef long long ll;
using namespace std;
const int maxn = 2005;

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 rank1(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(2005);
    int cas=1;
    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 = rank1(A, maxp+1, n);
        //printf("Case #%d:\n%lld", cas++,(1ll << (n-r)) - 1);
        long long ans=1;
        for(int i=1;i<=n-r;i++)
        {
            ans*=2;
            ans%=1000000007;
        }
        printf("Case #%d:\n%lld\n", cas++,ans-1);
    }
    return 0;
}