Codeforces Round #232 (Div. 2) C

C. On Number of Decompositions into Multipliers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an integer m as a product of integers a1, a2, ... an . Your task is to find the number of distinct decompositions of number m into the product of n ordered positive integers.

Decomposition into n products, given in the input, must also be considered in the answer. As the answer can be very large, print it modulo1000000007 (109 + 7).

Input

The first line contains positive integer n (1 ≤ n ≤ 500). The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

In a single line print a single number k — the number of distinct decompositions of number m into n ordered multipliers modulo 1000000007(109 + 7).

input

1
15

output

1

分析：用map存储每个素数的个数接着就是组合公式c(n+k-1,k-1),因为是乘法所以相当于往盒子里面放小球盒子可以为空。因此多出n个盒子

 1 #include<cstring>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<map>
 5 typedef long long LL;
 6 using namespace std;
 7 const int MAX =;
 8 const int F = 1e6+;
 9 const int MOD = 1e9+;
 map<int , int > m;
 int a[MAX];
 LL c[][MAX];
 void getp(int n)
 {
     long long i;
     for(i=;(long long)i*i<=n;i++)
     {
         while(n%i==)
         {
             m[i]++;
             n/=i;
         }
     }
     if( n !=  ) m[n]++;
 }
 void init()
 {
     c[][]=;
     for(int i=;i<;i++)
     {
         c[i][i]=c[i][]=;
         for(int j=;j<=min(i,MAX);j++)
         {
             c[i][j]=(c[i-][j]+c[i-][j-])%MOD;
         }
     }
 }
 int main()
 {
     int n;
     LL ans;
     while(scanf("%d",&n)==)
     {
         m.clear(); ans=;
         for(int i=;i<n;i++)
         {
             scanf("%d",&a[i]);
             getp(a[i]);
         }
         init();
         //printf("s");
         for(map<int,int> ::iterator it=m.begin();it!=m.end();it++)
         {
             int k=it->second;
             ans=ans*c[k-+n][n-]%MOD;
         }
         printf("%I64d\n",ans);
     }
     return ;

60 }