BZOJ2721 樱花 [线性筛]

2721: [Violet 5]樱花

Time Limit: 5 Sec  Memory Limit: 128 MB
Submit: 646  Solved: 378
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Description

Input

Output

题意就是求有多少正整数数对(x,y)满足

1x+1y=1n!

由于x,y>0，故显然有y>n!
不妨设y=n!+t(t>0)，那么有
1x+1n!+t=1n!
化简后得到
n!(n!+t)+x(n!)=x(n!+t)
x=(n!)2t+n!
故答案为d((n!)2)
我们令

y=n!+t(t>0)

=>1x+1n!+t=1n!

那么

=>x(n!)+n!(n!+t)=x(n!+t)

=>x=(n!)2t+n!=>(n!)∗(2t+1)

#include<bits/stdc++.h>using namespace std;const int N = 1001005,mod = 1000000007;int primes,prime[N],n,m;long long ans=1,tmp,ret;bool vis[N];void init(){    for(register int i=2;i<=n;i++){        if(!vis[i])prime[++primes]=i;        for(register int j=1;prime[j]*i<=n;j++){            vis[prime[j]*i]=1;            if(i%prime[j]==0)break;        }    }}int main(){    scanf("%d",&n);    init();    for(register int i=1;i<=primes;i++){        tmp=n,ret=0;        while(tmp)ret+=tmp/prime[i],tmp/=prime[i];        ret=(ret<<1|1)%mod;        ans*=ret;ans%=mod;    }    printf("%lld\n",ans);    return 0;}