HDU 4390 Number Sequence(容斥原理)

Number Sequence

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 822    Accepted Submission(s): 339

Problem Description

Given a number sequence b₁,b₂…b_n.
Please count how many number sequences a₁,a₂,...,a_n satisfy the condition that a₁*a₂*...*a_n=b₁*b₂*…*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 b₁, b₂,...,b_n(1<b_i<=1000000, b₁*b₂*…*b_n<=10²⁵).

 

Output

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

 

Sample Input

2

3 4

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<queue>
#include<cmath>
using namespace std;

#define mod 1000000007
#define LL long long
int prim[800000], f[1000000];
int c[505][505];
vector<int> V;
int len = 0, n;
/**首先是将所有 的b进行素因子分解，则a和b的因子是完全一致的。
剩下的便是将所有b的因子，分给a
我们考虑某个素因子pi,如果有ci个，便成了子问题将ci个相同的物品放入到n个不同的容器中，种数为多少
但是要求ai>1，也就是容器不能为空，这是个问题。
我们考虑的是什么的情况，然后减去假设有一个确定是空的情况，发现可以用容斥原理解决
我们假设f[i]表示有i个容器的结果，c(n,i)*f[i]
将m个物品放到到不同的n个容器中，结果为c(n+m-1,n-1)**/
void init(int n) {
    prim[0] = 2; len = 1;
    for(int i = 3; i < n; i += 2) if(!f[i]) {
        prim[len++] = i;
        for(int j = 3 * i; j < n; j += 2*i) f[j] = true;
    }
    for(int i = 0; i <= 500; i++){
        c[i][0] = c[i][i] = 1;
        for(int j = 1; j < i; j++)
            c[i][j] = (c[i-1][j-1] + c[i-1][j]) % mod;
    }
}
void resolve(int d){
    for(int i = 0; prim[i] <= d; i++){
        while(d % prim[i] == 0){ V.push_back(prim[i]); d /= prim[i]; }
    }
    //for(int i = 2; i*i <= d; i++){
       // while(d % i == 0){ d /= i; V.push_back(i);}
    //}
    if(d > 1)V.push_back(d);
}
LL get(int n, int m){
    return c[n+m-1][n-1];
}
LL solve(){
    sort(V.begin(), V.end());
    int a[1020] = {1}, l = 0;
    for(int i = 1; i < V.size(); i++){
        if(V[i] != V[i-1])a[++l] = 1;
        else a[l] ++;
    }
    LL ans = 1;
    for(int i = 0; i <= l; i++)ans = (ans * get(n,a[i])) % mod;
    for(int i = 1; i < n; i++){
        LL t = c[n][i];//????
        for(int j = 0; j <= l; j++){
            //n-i相当于有i个空位
            t = (t * get(n-i, a[j])) % mod;
        }
        if( i&1 )ans = ((ans - t) % mod + mod) % mod;
        else ans = (ans + t) % mod;
    }
    return ans;
}
int main()
{
    int i, d;
    init(1000000);
    while(~scanf("%d", &n)){
        V.clear();//初始化
        for(i = 0; i < n; i++){
            scanf("%d", &d);
            resolve(d);
        }
        cout<<solve()<<endl;
    }
    return 0;
}