hdu4390容斥定理加组合数

Given a number sequence b 1,b 2…b n. 
Please count how many number sequences a 1,a2,...,a n satisfy the condition that a 1*a 2*...*an=b 1*b 2*…*b n (a i>1).

Input

The input consists of multiple test cases. 
For each test case, the first line contains an integer n(1<=n<=20). The second line contains n integers which indicate b 1, b 2,...,b n(1<bi<=1000000, b 1*b 2*…*b n<=10 25).

Output

For each test case, please print the answer module 1e9 + 7.

Sample Input

2
3 4

Sample Output

4

        
  

Hint

For the sample input, P=3*4=12.
Here are the number sequences that satisfy the condition:
2 6
3 4
4 3
6 2

        
 

#include<iostream>>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 1005
#define ll long long
const ll mod=1e9+7;
#define maxm 1000005
using namespace std;
int p[maxm];
int a[maxm];
ll c[maxn][maxn];
int vis[maxm];
void init()
{
    int i,j;
    for(i=0;i<maxn;i++)
    {
        c[i][i]=c[i][0]=1;
        for(j=1;j<i;j++)
            c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;
    }
}
ll getnum(int m,int n)
{
    return c[m+n-1][n-1];
}
ll solve(int cnt,int n)
{
    int i,j;
    ll ans=1;
    for(i=0;i<=cnt;i++)
        ans=(ans*getnum(a[i],n))%mod;
    for(i=1;i<n;i++)
    {
        ll temp=c[n][i];
        for(j=0;j<=cnt;j++)
            temp=(temp*getnum(a[j],n-i))%mod;
        if(i&1)
            ans=(ans-temp+mod)%mod;
            else
                ans=(ans%mod+temp%mod)%mod;

    }
    return ans;
}
int main()
{
    init();
    int n;
    while(~scanf("%d",&n))
    {

        int t,i,j;
        int cnt=0;
        for( i=0;i<n;i++)
        {
            scanf("%d",&t);
          for( j=2;j*j<=t;j++)
          {
              while(t%j==0)
              {
                  p[cnt++]=j;
                  t/=j;
              }
          }
          if(t>1)
            p[cnt++]=t;
        }
       sort(p,p+cnt);
       a[0]=1;
       i=0;
       for(j=1;j<cnt;j++)
       {
           if(p[j]==p[j-1])
            a[i]++;
           else
            a[++i]=1;
       }
       cnt=i;
       printf("%lld\n",solve(cnt,n));
    }
    return 0;
}