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Description$Description$

求出有多少對 (x,y)$(x,y)$ 滿足

Solution$Solution$

首先是對公式進行變形

這下問題就轉化成求

我們寫出 n!$n!$ 的唯一分解式：

那麼 (n!)2$(n!{)}^2$ 的約數個數等於

Code$Code$

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

#define N 10000000
#define P 1000000007
int prime[N], number[N], f[N];
int main() {
    int n; int cnt = 0;
    scanf("%d", &n);
    for (int i = 2; i <= n; i++) {    // 線性篩
        if (f[i] == 0) {
            cnt++;
            prime[cnt] = i; 
            f[i] = cnt;
        }
        for (int j = 1; j <= cnt; j++){
            if
 
 (prime[j] * i > n) break;
            f[prime[j] * i] = j;
            if (i % prime[j] == 0) break;
        }
    }   
    for (int i = 2; i <= n; i++) {
        int x = i;
        while (x != 1) {        // 分解 i
            int j = prime[f[x]];
            while (x % j == 0) {
                number[f[x]]++;
                x /= j;
            }
        }
    }
    long long sum = 1;
    for (int i = 1; i <= cnt; i++)   // 計算結果
        if (number[i] != 0) 
            (sum *= (2ll * number[i] + 1ll)) %= P;
    printf("%lld\n", sum);
    return 0;
}