HDU5833（2016CCPC网赛）——Zhu and 772002（异或方程组，素数分解）

Zhu and 772002

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 510    Accepted Submission(s): 170

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

 

完全平方数就是每一个素因子的个数都是偶数

那么我们把每个数进行质因数分解，奇数个记为1，偶数个记为0，最后就是求一个异或方程组的解的个数。

#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
using namespace std;
const int MAXN =310;
const int INF =1000000007 ;
const int MOD =1000000007;
const double EPS=1e-8;
const double pi = acos(-1);
int prime[MAXN];
int A[MAXN][MAXN];
int num=0;
bool is_prime[2010]={true};
long long powmod(long long a,long long n)//快速幂取膜
{
    long long res=1;
    while(n){
        if(n&1) res=res*a%MOD;
        a=a*a%MOD;
        n>>=1;
    }
    return res;
}

long long solve(int m,int n)//根据系数矩阵求自由变元数量
{
    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++;
    }
    long long ans=powmod(2,n-i)-1;
    return ans;
}

long long a[MAXN];
void getprime()//求出2000以内素数表
{
    memset(is_prime,true,sizeof(is_prime));
    is_prime[2]=true;
    for(int i=2;i<=2000;i++){
        if(is_prime[i]){
            prime[num++]=i;
            for(int j=2*i;j<=2000;j+=i){
                is_prime[j]=false;
            }
        }
    }
}
void get_key(int i)//素数分解
{
    long long temp=a[i];
    for(int k=0;k<num;k++){
        while(temp%prime[k]==0){
            A[k][i]^=1;
            temp/=prime[k];
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    getprime();
    for(int cas =1;cas<=t;cas++){
        int n;
        scanf("%d",&n);
        memset(A,0,sizeof(A));
        printf("Case #%d:\n",cas);
        for(int i=0;i<n;i++){
            scanf("%I64d",a+i);
            get_key(i);
        }
        printf("%I64d\n",solve(num,n));
    }
}