Factors of Factorial（AtCoder-2286）

Problem Description

You are given an integer N. Find the number of the positive divisors of N!, modulo 109+7.

Constraints

• 1≤N≤103

Input

The input is given from Standard Input in the following format:

N

Output

Print the number of the positive divisors of N!, modulo 109+7.

Example

Sample Input 1

3

Sample Output 1

4
There are four divisors of 3! =6: 1, 2, 3 and 6. Thus, the output should be 4.

Sample Input 2

6

Sample Output 2

30

Sample Input 3

1000

Sample Output 3

972926972

题意：给出一个整数 n，求 n! 的约数个数

思路：

根据唯一分解定理，一个整数 n 可分解为：

那么，n 的约数个数为：

故而对于每个素因子，可以选择 0~ai 个，根据乘法原理，直接统计即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 1000+5;
const int dx[] = {0,0,-1,1,-1,-1,1,1};
const int dy[] = {-1,1,0,0,-1,1,-1,1};
using namespace std;
int num[N];
int main() {
    int n;
    scanf("%d",&n);
    for(int i=2;i<=n;i++){
        int temp=i;
        for(int j=2;j<=temp;j++){
            while(temp%j==0){
                num[j]++;
                temp/=j;
            }
        }
        if(temp!=1)
            num[temp]++;
    }

    LL res=1;
    for(int i=1; i<=n; i++)
        if(num[i])
            res=(res*(num[i]+1))%MOD;
    printf("%lld\n",res);

    return 0;
}