Factors of Factorial

问题 G: Factors of Factorial

时间限制: 1 Sec  内存限制: 128 MB
提交: 45  解决: 25
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题目描述

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

Constraints
1≤N≤103

输入

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

输出

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

样例输入

3

样例输出

4

提示

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

求n的阶乘的因子个数

将这个数表示成质因子的乘积

n!= 2^a + 3^b + 5^c + 7^d……

因子个数就是（a+1）*（b+1）*(c+1)*……

即对2来说可以选择0到a个

代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MOD=1e9+7;
const int INF=1e5+100;
int f[INF];
int main()
{
    int n,a;
    cin>>n;
    memset(f,0,sizeof(f));
    for(int i=2; i<=n; i++){
        a=i;
        for(int j=2; j<=a; j++){
            while(a%j==0){
                f[j]++;
                a=a/j;
            }
        }
        if(a!=1)  f[a]++;
    }
    ll ans=1;
    for(int i=1; i<=n; i++){
        if(f[i]!=0) ans=(ans*(f[i]+1))%MOD;
    }
    cout<<ans%MOD<<endl;
    return 0;
}