正题

题目链接:https://www.luogu.org/problemnew/show/P1445

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题目大意

求有多少个$x,y$满足
$$\frac{1}{x}+\frac{1}{y}=\frac{1}{n!}$$

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解题思路

$$\frac{x+y}{xy}=\frac{1}{n!}$$
$$n!(x+y)=xy$$
$$xy-n!(x+y)=0$$
$$xy-n!(x+y)+(n!{)}^2=(n!{)}^2$$
$$(x-n!)(y-n!)=(n!{)}^2$$
然后设$a=x-n!,b=y-n!$
所以求有多少个$a$和$b$就对了
然后$$n!=\prod p_i^{c_i}$$
$$(n!{)}^2=\prod p_i^{2\ast c_i}$$
然后计算$a$个数就可以推出$b$了
答案就是$$\prod (2\ast c_i+1)$$

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$code$

#include<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
const ll XJQ=1e9+7,N=1e6+10;
ll n,ans=1,pri[N],cnt,c[N];
bool v[N];
void Prime(ll x)
{for(ll i=2;i<=x;i++){if(v[i]) continue;pri[++cnt]=i;for(ll j=i;j<=x;j+=i)v[j]=1;}
}
void Primes(ll n)
{ll l=0;for(ll i=1;i<=cnt;i++){if(pri[i]*pri[i]>n) break;if(n%pri[i]) continue;while(n%pri[i]==0)n/=pri[i],c[i]++;}if(n==1) return;int mark=lower_bound(pri+1,pri+1+cnt,n)-pri;c[mark]++;
}
int main()
{scanf("%lld",&n);Prime(n);for(ll i=2;i<=n;i++)Primes(i);for(ll i=1;i<=cnt;i++)(ans*=c[i]*2+1)%=XJQ;printf("%lld",ans);
}