【BZOJ-2721】樱花 线性筛 + 数学

2721: [Violet 5]樱花

Time Limit: 5 Sec  Memory Limit: 128 MB
Submit: 499  Solved: 293
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Description

Input

Output

Sample Input

 3

Sample Output

 9

HINT

Source

interviewstreet--EQUATIONS

Solution

巧妙！

$frac{1}{x}+frac{1}{y}=frac{1}{n!}$ 令$z=n!$

则可以得到$frac{1}{x}+frac{1}{y}=frac{1}{z}=>x=frac{yz}{y-z}$

再另$t=y-z$则可以得到$x=z+frac{z^{2}}{t}$

所以我们求$n!^{2}$的约数，就是答案，这就利用到线筛

Code

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
#define P 1000000007
#define LL long long
int N,cnt,prime[1000100],z[1000100],tmp;
bool flag[1000100];
LL ans=1LL;
void Getprime()
{
    flag[1]=1; cnt=0;
    for (int i=2; i<=N; i++)
        {
            if (!flag[i]) prime[++cnt]=i;
            for (int j=1; j<=cnt && prime[j]*i<=N; j++)
                {
                    flag[i*prime[j]]=1;
                    if (prime[j]%i==0) break;
                }
        }
}
void Calc(int x)
{
    for (int i=prime[x]; i<=N; i+=prime[x])
        for (int j=i; j%prime[x]==0; j/=prime[x]) z[x]++;
}
int main()
{
    scanf("%d",&N);
    Getprime();
    for (int i=1; i<=cnt; i++) Calc(i);
    for (int i=1; i<=cnt; i++) printf("%d   ",z[i]); puts("");
    for (int i=1; i<=cnt; i++) ans=((LL)ans*(z[i]<<1|1)%P)%P;
    printf("%lld
",ans);
    return 0;
}