HDU 5833 Zhu and 772002 2016中国大学生程序设计竞赛 - 网络选拔赛（高斯消元）

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中国大学生程序设计竞赛 - 网络选拔赛

Recommend

wange2014 | We have carefully selected several similar problems for you: 5842 5841 5840 5839 5838

原题醉了。。CCPC网络赛居然出原题。。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;
}