[bzoj2721][数论]樱花

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Description

Input

Output

Sample Input

2

Sample Output

3

HINT

题解

设 $z=n!$
化简可得
$\frac{zx}{x-z}=y$
显然 $x,y$均大于 $z$
不妨设 $x=z+d$
写为
$\frac{z^2}{d}+d=y$
由于 $y$是整数，显然求 $z^2$的因数个数
大力线性筛

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<ctime>
#include<map>
#define LL long long
#define mp(x,y) make_pair(x,y)
#define mod 1000000007
using namespace std;
inline int read()
{
	int f=1,x=0;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
inline void write(int x)
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
}
inline void print(int x){write(x);printf(" ");}
int pr[1100000],ok[1100000],plen,n;
bool v[1100000];
void getpri(int MAXN)
{
	memset(v,true,sizeof(v));ok[1]=1;
	for(int i=2;i<=MAXN;i++)
	{
		if(v[i])pr[++plen]=i,ok[i]=i;
		for(int j=1;j<=plen&&i*pr[j]<=MAXN;j++)
		{
			v[i*pr[j]]=false;ok[i*pr[j]]=min(ok[i],ok[pr[j]]);
			if(!(i%pr[j]))break;
		}
	}
}
LL cnt[1100000];
void calc()
{
	for(int i=1;i<=n;i++)
	{
		int tmp=i;
		while(tmp!=1)
		{
			cnt[ok[tmp]]++;
			tmp/=ok[tmp];
		}
	}
}
int main()
{
	getpri(1000000);
	n=read();
	calc();
	LL ans=1;
	for(int i=1;i<=n;i++)if(v[i])ans=ans*(2*cnt[i]+1)%mod;
	printf("%lld\n",ans);
	return 0;
}