• 洛谷P1829 Crash的数字表格【莫比乌斯反演】


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    题意

    计算 (sum_{i=1}^{n}sum_{j=1}^{m}lcm(i,j))

    题解

    [egin{aligned}Ans=sum_{i=1}^{N}sum_{j=1}^{M}lcm(i,j)&=sum_{i=1}^{N}sum_{j=1}^{M}frac{ij}{gcd(i,j)}=sum_{d=1}^{N}sum_{i=1}^{N}sum_{j=1}^{M}[gcd(i,j)=d]frac{ij}{d}\&=sum_{d=1}^{N}frac{1}{d}sum_{i=1}^{N}sum_{j=1}^{M}[gcd(i,j)=d]ij\end{aligned} ]

    设:

    [f(d)=sum_{i=1}^{N}sum_{j=1}^{M}[gcd(i,j)=d]ij\F(n)=sum_{n|d}f(d)=sum_{n|i}^{N}sum_{n|j}^{M}ij=sum_{i=1}^{frac{N}{n}}sum_{j=1}^{frac{M}{n}}incdot jn=n^2sum_{i=1}^{frac{N}{n}}sum_{j=1}^{frac{M}{n}}ij ]

    然后反演 (f(n))

    [egin{aligned}f(n)&=sum_{n|d}mu(frac{d}{n})F(d)=sum_{n|d}mu(frac{d}{n})d^2sum_{i=1}^{frac{N}{d}}sum_{j=1}^{frac{M}{d}}ij=sum_{t=1}^{frac{N}{n}}mu(t)n^2t^2sum_{i=1}^{frac{N}{nt}}sum_{j=1}^{frac{M}{nt}}ij\&=n^2sum_{t=1}^{frac{N}{n}}mu(t)t^2sum_{i=1}^{frac{N}{nt}}isum_{j=1}^{frac{M}{nt}}jend{aligned} ]

    然后代回去求 (Ans)

    [egin{aligned}Ans=sum_{n=1}^{N}frac{1}{n}f(n)&=sum_{n=1}^{N}nsum_{t=1}^{frac{N}{n}}mu(t)t^2sum_{i=1}^{frac{N}{nt}}isum_{j=1}^{frac{M}{nt}}j=sum_{T=1}^{N}Tsum_{t|T}mu(t)tsum_{i=1}^{frac{N}{T}}isum_{j=1}^{frac{M}{T}}jend{aligned} ]

    关于 (f(T)=sum_{t|T}mu(t)t),可以证明它是一个积性函数:

    [T(mu*id_{-1})(T)=Tsum_{t|T}mu(t)frac{t}{T}=sum_{t|T}mu(t)t ]

    (n,p) 互质,那么满足积性函数要求, (f(np)=f(n)f(p))
    (p|n),那么 (f(np)=f(n)),因为 (np)(n) 多的因数必然有多个因子 (p),所以这部分因数的 (mu) 值为 (0)

    代码

    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #include <stdlib.h>
    using namespace std;
    typedef long long LL;
    const int N=1e7+10;
    const int mod=20101009;
    int n,m,vis[N],prime[N],cnt;
    LL sum[N],f[N];
    
    void init(int n){
    	f[1]=1;
    	for(int i=2;i<=n;i++){
    		if(!vis[i]) {prime[++cnt]=i;f[i]=(1-i+mod)%mod;}
    		for(int j=1;1ll*i*prime[j]<=n;j++){
    			vis[i*prime[j]]=1;
    			if(i%prime[j]==0) {f[i*prime[j]]=f[i];break;}
    			f[i*prime[j]]=f[i]*f[prime[j]]%mod;
    		}
    	}
    	for(int i=1;i<=n;i++) sum[i]=(f[i]*i+sum[i-1])%mod;
    }
    
    int main(){
    	init(1e7);
    	scanf("%d%d",&n,&m);
    	if(n>m) swap(n,m);
    	LL ans=0;
    	for(int l=1,r;l<=n;l=r+1){
    		r=min(n/(n/l),m/(m/l));
    		LL temp1=1ll*(n/l+1)*(n/l)/2%mod;
    		LL temp2=1ll*(m/l+1)*(m/l)/2%mod;
    		ans=(ans+(sum[r]-sum[l-1]+mod)%mod*temp1%mod*temp2%mod)%mod;
    	}
    	printf("%lld
    ",ans);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/BakaCirno/p/12530943.html
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