#莫比乌斯反演#洛谷 5221 Product

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#洛谷 5221 Product

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题目

求 $∏i=1n∏j=1nlcm(i,j)gcd(i,j)\prod_{i=1}^n\prod_{j=1}^n\frac{lcm(i,j)}{gcd(i,j)}$

分析

又是一波推式子的过程
$=∏i=1n∏j=1nij1gcd(i,j)2=\prod_{i=1}^n\prod_{j=1}^nij\frac{1}{gcd(i,j)^2}$
$=(n!)2n(∏i=1n∏j=1ngcd(i,j))−2=(n!)^{2n}(\prod_{i=1}^n\prod_{j=1}^ngcd(i,j))^{-2}$
$=(n!)2n(∏d=1n∏i=1n∏j=1n[gcd(i,j)==d])=(n!)^{2n}(\prod_{d=1}^n\prod_{i=1}^n\prod_{j=1}^n[gcd(i,j)==d])$
$=(n!)2n(∏d=1nd∏i=1⌊nd⌋∏j=1⌊nd⌋[gcd(i,j)==1])=(n!)^{2n}(\prod_{d=1}^nd\prod_{i=1}^{\lfloor\frac{n}{d}\rfloor}\prod_{j=1}^{\lfloor\frac{n}{d}\rfloor}[gcd(i,j)==1])$
$=(n!)2n(∏d=1nd∑i=1⌊nd⌋∑j=1⌊nd⌋[gcd(i,j)==1])\large =(n!)^{2n}(\prod_{d=1}^nd^{\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}[gcd(i,j)==1]})$
指数就是 $2∗φ(⌊nd⌋)−12*\varphi(\lfloor\frac{n}{d}\rfloor)-1$
就可以O(n)求解了

代码

#include <cstdio>
#define rr register
using namespace std;
const int mod=104857601,N=1000101;
int n,phi[N],prime[N],cnt;
inline signed ksm(int x,int y){
	rr int ans=1;
	for (;y;y>>=1,x=1ll*x*x%mod)
	    if (y&1) ans=1ll*ans*x%mod;
	return ans;
}
signed main(){
	scanf("%d",&n); int ans=1,ans1=1;
	for (rr int i=1;i<=n;++i) phi[i]=i;
	for (rr int i=2;i<=n;++i){
		if (phi[i]==i) prime[++cnt]=i,phi[i]=i-1;
		for (rr int j=1;j<=cnt&&prime[j]*i<=n;++j){
			if (i%prime[j]==0){
				phi[i*prime[j]]=phi[i]*prime[j];
				break;
			}
			phi[i*prime[j]]=phi[i]*(prime[j]-1);
		}
		ans=1ll*ans*i%mod;
	}
	for (rr int i=1;i<=n;++i) phi[i]=(phi[i-1]+phi[i]*2)%(mod-1); ans=ksm(ans,n<<1);
	for (rr int i=2;i<=n;++i) ans1=1ll*ans1*ksm(i,phi[n/i]-1)%mod;
	return !printf("%d",1ll*ans*ksm(ans1,mod-3)%mod);	
}