bzoj2721 [Violet 5]樱花

Description

Input

Output

 

Sample Input

 

Sample Output

 

HINT

 

 

蛋疼的推公式题……依题意1/x+1/y=1/z，令y=z+d，然后

1/x+1/(z+d)=1/z

(x+z+d)/(xz+xd)=1/z

xz+z^2+dz=xz+xd

z^2+dz=xd

x=z^2/d+z

显然x是正整数主要取决于d能整除z^2

每一个d都对应一个唯一的x，所以答案就是z^2的约数个数

即（n!）^2的约数个数

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define mod 1000000007
#define mx 1000000
using namespace std;
inline LL read()
{
    LL x=0,f=1;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(LL a)
{
    if (a<0){printf("-");a=-a;}
    if (a>=10)write(a/10);
    putchar(a%10+'0');
}
inline void writeln(LL a){write(a);printf("\n");}
bool prime[mx+10];
LL rep[mx+10];
int n;
LL ans=1;
inline void shai()
{
    memset(prime,1,sizeof(prime));
    prime[1]=0;
    for (int i=2;i<=n;i++)
    if (prime[i])
    {
        LL res=i;
        while ((LL)n/res>0)
        {
            rep[i]+=(LL)n/res;
            res*=i;
        }
        for (int j=2*i;j<=n;j+=i)
            prime[j]=0;
    }
}
int main()
{
    n=read();shai();
    if (n==1)
    {
        printf("1\n");
        return 0;
    }
    for (int i=1;i<=n;i++)
        if (prime[i])
        {
            ans=ans*(2*rep[i]+1)%mod;
        }
    printf("%lld\n",ans);
}