樱花

https://www.luogu.org/problem/P1445

很容易得$x=frac{n! 	imes y}{y-n!} $

设$k=y-n!$
则原式x=n!(k+n!)/k
      x=n!^2+n!k/k
      x=n!^2/k+n!
      ∴只有k|n^2时x才为正数
      ∴k的个数即是x、y的个数
      易看出k的个数即n!^2约数的个数
      于是这道题转换为求n^2的约数的个数有多少
      公式:百度百科自己搜

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define MAXN 1000050
#define int long long
#define N 1000000
using namespace std;
const int mod=1e9+7;
inline int read(){
	int x=0,f=1;
	char ch=getchar();
	while('0'>ch || ch>'9'){if(ch=='-') f=-1; ch=getchar();}
	while('0'<=ch && ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	return x*f;
}
int n,pre[MAXN],book[MAXN],cnt,res,f[MAXN],ans=1;
void shai(){
	rep(i,2,N){
		if(!book[i]){
			book[i]=i; pre[++cnt]=i;
		}
		rep(j,1,cnt){
			if(i*pre[j]>N || pre[j]>book[i]) break;
			book[i*pre[j]]=pre[j];
		}
	}
}
main(){
	n=read();
	shai();
	rep(i,2,n){
		int x=i;
		while(x!=1)f[book[x]]++,x/=book[x];
	}
	rep(i,2,n) f[i]*=2;
	rep(i,1,n) ans=(ans*(f[i]+1))%mod;
	printf("%lld",ans);
	return 0;
}