#莫比乌斯反演#洛谷 5176 公约数

题目

$$\sum _{i=1}^n\sum _{j=1}^m\sum _{k=1}^{p}gcd(ij,ik,jk)\times gcd(i,j,k)\times (\frac{gcd(i,j)}{gcd(i,k)gcd(j,k)}+\frac{gcd(i,k)}{gcd(i,j)gcd(j,k)}+\frac{gcd(j,k)}{gcd(i,j)gcd(i,k)})$$

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分析

$$\because gcd(ij,ik,jk)=\frac{gcd(i,j)gcd(i,k)gcd(j,k)}{gcd(i,j,k)}$$
再加上后面的通分，这个式子可以火速化简成
$$\sum _{i=1}^n\sum _{j=1}^m\sum _{k=1}^{p}gcd(i,j{)}^2+gcd(i,k{)}^2+gcd(j,k{)}^2$$
$$=p\times \sum _{i=1}^n\sum _{j=1}^{m}gcd(i,j{)}^2+m\times \sum _{i=1}^n\sum _{k=1}^{p}gcd(i,k{)}^2+n\times \sum _{j=1}^m\sum _{k=1}^{p}gcd(j,k{)}^2$$
于是题目就变成了这个洛谷 4449
对于单独的$$\sum _{i=1}^n\sum _{j=1}^{m}gcd(i,j{)}^2$$
最后变成$$\sum _{i=1}^n\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{i}\rfloor \sum _{d|i}{d}^2\mu (\frac{i}{d})$$
至于后面怎么推就理解吧

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代码

#include 
#include 
#define rr register
using namespace std;
const int N=20000011,mod=1000000007;
int dp[N],prime[N],cnt,T,k; bool v[N];
inline signed iut(){
	rr int ans=0; rr char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans;
}
inline void print(int ans){
	if (ans>9) print(ans/10);
	putchar(ans%10+48);
}
inline signed min(int a,int b){return a<b?a:b;}
inline signed mo(int x,int y){return x+y>=mod?x+y-mod:x+y;}
inline void init(){
	dp[1]=1;
	for (rr int i=2;i<N;++i){
		if (!v[i]) prime[++cnt]=i,dp[i]=1ll*(i+1)*(i-1)%mod;
		for (rr int j=1;prime[j]*i<N&&j<=cnt;++j){
			v[i*prime[j]]=1;
			if (i%prime[j]==0){
				dp[i*prime[j]]=1ll*dp[i]*(dp[prime[j]]+1)%mod;
				break;
			}
			dp[i*prime[j]]=1ll*dp[i]*dp[prime[j]]%mod;
		}
	}
	for (rr int i=2;i<N;++i) dp[i]=mo(dp[i],dp[i-1]);
}
inline signed f(int n,int m){
	rr int t=n<m?n:m,ans=0;
	for (rr int l=1,r;l<=t;l=r+1){
		r=min(n/(n/l),m/(m/l));
		ans=mo(ans,1ll*mo(dp[r]-dp[l-1],mod)*(n/l)%mod*(m/l)%mod);
	}
	return ans;	
}
signed main(){
	T=iut(); init();
	while (T--){
		rr int n=iut(),m=iut(),p=iut(),ans=0;
		ans=mo(mo(1ll*n*f(m,p)%mod,1ll*m*f(n,p)%mod),1ll*p*f(n,m)%mod);
		print(ans),putchar(10);
	}
	return 0;
}