Number Sequence

2
3 4

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<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define N 100010
#define mod 1000000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos(-1.0);
const double E=2.718281828;
typedef long long ll;

const int INF=1000010;

using namespace std;

int C[1001][45];///组合数
map<int,int>mp;

void init()
{
    for(int i=0; i<=1000; i++)
        C[i][0]=1;///i个选0个为1；
    for(int i=1; i<=1000; i++)
        for(int j=1; j<=42; j++)
            C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;///c(n,k)=c(n-1,k)+c(n-1,k-1),可以自己证明
}

void Fenjie(int n)///分解质因数
{
    for(int i=2; i*i<=n; i++)
    {
        if(n%i==0)
        {
            while(n%i==0)
            {
                mp[i]++;
                n/=i;
            }
        }
        if(n==1)
            break;
    }
    if(n>1)
        mp[n]++;
}

int main()
{
    init();
    int n;
    while(cin>>n)
    {
        mp.clear();
        int b;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&b);
            Fenjie(b);
        }
        ll ans=0;
        for(int i=0; i<n; i++)///容斥原理
        {
            ll tem=C[n][i];
            map<int,int>::iterator it;
            for(it=mp.begin(); it!=mp.end(); it++)///计算有i个盒子为空的组合数
            {
                int t=it->second+n-i-1;
                tem*=C[t][n-i-1];
                tem%=mod;
            }
            if(i%2)
                ans-=tem,ans+=mod;
            else
                ans+=tem;
            ans%=mod;
        }
        cout<<ans%mod<<endl;
    }
    return 0;
}