HDU 5833 Zhu and 772002 （高斯消元）

Zhu and 772002

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5833

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

Source

2016中国大学生程序设计竞赛 - 网络选拔赛


##题意：

给出n个数，求有多少种方式使得选取的数的乘积是一个完全平方数.


##题解：

原题：[UVA11542](http://acm.hust.edu.cn/vjudge/problem/34393) (大白书P160例题25)
转化成异或方程组，并用高斯消元求解矩阵的秩.
很遗憾，上述知识点都不会....


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 2100
#define mod 1000000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
typedef long long ll;
using namespace std;

typedef int Matrix[maxn][maxn];

int prime[maxn], vis[maxn];

Matrix A;

int get_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 gauss(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;

}

LL quickmod(LL a,LL b,LL m)

{

LL ans = 1;

while(b){

if(b&1){

ans = (ansa)%m;

b--;

}

b/=2;

a = aa%m;

}

return ans;

}

int main() {

//freopen("in.txt", "r", stdin);
int m = get_primes(2100);

int t;
int ca = 1;
scanf("%d", &t);
while (t--) {
    printf("Case #%d:\n", ca++);
    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 = gauss(A, maxp+1, n);
    LL ans = quickmod(2, (LL)(n-r), mod) - 1;
    //printf("%lld\n", (1LL << (n-r)) - 1);
    printf("%lld\n", ans);
}
return 0;

}