#莫比乌斯反演#BZOJ 2694 LCM

题目

多组询问求

\[\sum_{i=1}^n\sum_{j=1}^m{|\mu(\gcd(i,j))|*lcm(i,j)}\pmod {2^{30}} \]

\(T\leq 10^4,n,m\leq 4*10^6\)

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分析

设 \(f(n,m)=C(n+1,2)*C(m+1,2)\) ，
则根据Crash的数字表格类似的方法可以得到

\[\large \sum_{d=1}^{\min\{n,m\}}d\mu^2(d)\sum_{t=1}^{\min\{\lfloor\frac{n}{d}\rfloor,\lfloor\frac{m}{d}\rfloor\}}\mu(t)t^2*f(\lfloor\frac{n}{td}\rfloor,\lfloor\frac{m}{td}\rfloor) \]

这样整除分块只能做到 \(O(\min\{n,m\})\)，多组数据就不行了，
考虑维护 \(i=t*d\)，则

\[\large =\sum_{i=1}^{\min\{n,m\}}f(\lfloor\frac{n}{i}\rfloor,\lfloor\frac{m}{i}\rfloor)*(i\sum_{d|i}\mu(d)d\mu^2(\frac{i}{d})) \]

考虑 \(g(n)=\sum_{d|i}\mu(d)d\mu^2(\frac{i}{d})\) 是一个积性函数，

线性筛时只要质因子次数大于2时\(g(n)=0\)，

否则若质因数次数等于2则\(g(n*p)=-g(n)*p\)

那就可以做到\(O(T\sqrt{\min\{n,m\}})\)了

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代码

#include <cstdio>
#include <cctype>
#define rr register
using namespace std;
const int N=10000000;
const int mod=100000009;
typedef long long lll;
int f[N|15],prime[N|15],v[N|15],Cnt;
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 lll answ(lll n,lll m){return (n*(n+1)>>1)%mod*((m*(m+1)>>1)%mod)%mod;}
inline signed mo(int x,int y){return x+y>=mod?x+y-mod:x+y;}
signed main(){
	f[1]=1;
	for (rr int i=2;i<=N;++i){
		if (!v[i]) prime[++Cnt]=i,f[i]=mod-i+1;
		for (rr int j=1;j<=Cnt&&prime[j]<=N/i;++j){
			v[i*prime[j]]=1;
			if (i%prime[j]==0){
				f[i*prime[j]]=f[i];
				break;
			}
			f[i*prime[j]]=1ll*f[i]*f[prime[j]]%mod;
		}
	}
	for (rr int i=2;i<N;++i) f[i]=mo(f[i-1],1ll*f[i]*i%mod);
	for (rr int T=iut();T;--T){
		rr int n=iut(),m=iut(),ans=0;
		if (n>m) n^=m,m^=n,n^=m;
		for (rr int l=1,r;l<=n;l=r+1){
			r=min(n/(n/l),m/(m/l));
			ans=mo(ans,answ(n/l,m/l)*(f[r]-f[l-1]+mod)%mod);
		}
		print(ans),putchar(10);
	}
	return 0;
}