Luogu P5221 Product

題目
注意一下空間限制。
\(f(n)=\prod\limits_{i=1}^n\prod\limits_{j=1}^nij,g(n)=\prod\limits_{i=1}^n\prod\limits_{j=1}^n(i,j)\)
那麼答案就是\(f(n)g(n)^{-2}\).
顯然\(f(n)=(n!)^{2n}\)
\(g(n)=\prod\limits_{d=1}^nd^{\sum\limits_{i=1}^{\lfloor\frac nd\rfloor}\sum\limits_{j=1}^{\lfloor\frac nd\rfloor}[(i,j)=1]}=\prod\limits_{d=1}^nd^{\sum\limits_{i=1}^{\lfloor\frac nd\rfloor}2\varphi(i)-1}\)
而後就能夠直接作了。
注意一下因爲歐拉定理指數部分的要對\(P-1\)取模。c++

#include<bits/stdc++.h>
using namespace std;
const int N=1000007,P=104857601,M=104857600;
int pr[N>>3],m,phi[N];bool f[N];
int inc(int a,int b){return a+=b,a>=M? a-M:a;}
int dec(int a,int b){return a-=b,a<0? a+M:a;}
int mul(int a,int b){return 1ll*a*b%P;}
int sqr(int a){return mul(a,a);}
int power(int a,int k){int r=1;for(;k;k>>=1,a=mul(a,a))if(k&1)r=mul(a,r);return r;}
int fac(int n){int s=1;for(;n;--n)s=mul(s,n);return s;}
int inv(int a){return power(a,P-2);}
int cal1(int n){return power(fac(n),n<<1);}
int cal2(int n)
{
    phi[1]=f[1]=1;int ans=1;
    for(int i=2,j,x;i<=n;++i)
    {
    if(!f[i]) pr[++m]=i,phi[i]=i-1;
    for(j=1;j<=m&&i*pr[j]<=n;++j)
    {
        f[x=i*pr[j]]=1;
        if(i%pr[j]) phi[x]=phi[i]*phi[pr[j]];
        else {phi[x]=phi[i]*pr[j];break;}
    }
    }
    for(int i=2;i<=n;++i) phi[i]=inc(phi[i-1],phi[i]);
    for(int i=1;i<=n;++i) phi[i]=dec(inc(phi[i],phi[i]),1);
    for(int i=1;i<=n;++i) ans=mul(ans,power(i,phi[n/i]));
    return ans;
}
int main(){int n;cin>>n,cout<<mul(cal1(n),sqr(inv(cal2(n))));}
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