題意：

給出一個整數ｎ，求有多少個正整數對（x, y）滿足$\frac{1}{x}$ + $\frac{1} {y}$ = $\frac{1} {n!}$.答案對$10^9+7$取模

思路：

公式推導：

原式 =$\frac{x + y}{xy}$ =$\frac{1}{n!}$

$xn! + yn! = xy$

$xn! = xy - yn!$

$y = \frac{xn!}{x - n!}$

則，$y = \frac{xn!}{x - n!} = \frac{(x - n! + n!)n!}{x - n!} = n! + \frac{n! ^ 2}{x - n!}$

問題轉變為$n!^2$的約數個數

Code:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;

const int N = 1e6 + 10;
const int mod = 1e9 + 7;

int n;
int primes[N], cnt;
bool st[N];

void get(int n){
    for(int i = 2; i <= n; i ++){
        
 
if(!st[i]){
            primes[cnt ++] = i;
        }
        for(int j = 0; primes[j] <= n / i; j ++){
            st[i * primes[j]] = true;
            if(i % primes[j] == 0){
                break;
            }
        }
    }
}

int main(){
    cin.tie(0);
    cout.tie(0);

    get(N);

    cin >> n;

    
 
int ans = 1;
    for(int i = 0; i < cnt; i ++){
        int s = 0, p = primes[i], num = 1;
        while(n >= (ll)num * p){
            num *= p;
            s += n / num;
        }
        ans = (ll)ans * (2 * s + 1) % mod;
    }

    cout << ans << endl;

    return 0;
}