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  • 51Nod 1584 加权约数和

    51Nod 1584 加权约数和


    要求:(sum_{i=1}^nsum_{j=1}^nmax(i,j)sigma_1(ij))

    枚举(max(i,j)),式子变为:(2sum_{i=1}^nisum_{j=1}^isigma_1(ij)-sum_{i=1}^nisigma_1(i^2))

    这里不加证明地给出结论:(sigma_1^k(ij)=sum_{p|i}sum_{q|j}(pfrac{j}{q})^k[gcd(p,q)=1])

    先看右边部分,可以线性筛(sigma_1(n^2))

    (筛约数个数和的方法:(sigma_1(n=prod_{i=1}^mp_i^{e_i})=prod_{i=1}^m(1+p_i+cdots+p_i^{e_i})),筛平方的约数个数和类似,事实上任意(k)次方的都能筛)

    或者也可以推式子:

    (sum_{i=1}^nisigma_1(i^2))

    (=sum_{i=1}^nisum_{p|i}sum_{q|i}pfrac iqsum_{d|p,d|q}mu(d))

    (=sum_{i=1}^nisum_{p|i}sum_{q|i}pfrac iqsum_{d|p,d|q}mu(d))

    (=sum_{i=1}^nisum_{d|i}mu(d)sum_{d|p}^ipsum_{d|q}^ifrac iq)

    (=sum_{i=1}^nisum_{d|i}mu(d)sum_{p|frac id}^idpsum_{q|frac id}^ifrac i{dq})

    (=sum_{i=1}^nisum_{d|i}mu(d)dsigma_1(frac id)^2)

    左边部分类似,直接写结论了:

    (sum_{i=1}^nisum_{j=1}^isigma_1(ij)=sum_{i=1}isum_{d|i}mu(frac id)(frac id)sigma_1(d)S_{sigma_1}(d))

    其中(S_f(n)=sum_{i=1}^nf(i))


    右边用线性筛:

    #include<bits/stdc++.h>
    #define il inline
    #define vd void
    #define mod 1000000007
    typedef long long ll;
    il ll gi(){
    	ll x=0,f=1;
    	char ch=getchar();
    	while(!isdigit(ch))f^=ch=='-',ch=getchar();
    	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    	return f?x:-x;
    }
    int pr[1000010],P,sd[1000010],_sd[1000010],sum[1000010],ssd[1000010],mu[1000010];
    int sd2[1000010],_sd2[1000010],ssd2[1000010],sum2[1000010];
    bool yes[1000010];
    int ans[1000010];
    int main(){
    #ifdef XZZSB
    	freopen("in.in","r",stdin);
    	freopen("out.out","w",stdout);
    #endif
    	int N=1000000;
    	sd[1]=sd2[1]=1;mu[1]=1;
    	for(int i=2;i<=N;++i){
    		if(!yes[i])pr[++P]=i,sd[i]=sum[i]=i+1,_sd[i]=1,mu[i]=mod-1,sd2[i]=sum2[i]=(1+i+1ll*i*i)%mod,_sd2[i]=1;
    		for(int j=1;j<=P&&i*pr[j]<=N;++j){
    			yes[i*pr[j]]=1;
    			if(i%pr[j]==0){
    				sum[i*pr[j]]=(1ll*sum[i]*pr[j]+1)%mod;
    				sd[i*pr[j]]=1ll*_sd[i]*sum[i*pr[j]]%mod;
    				_sd[i*pr[j]]=_sd[i];
    				sum2[i*pr[j]]=(1ll*sum2[i]*pr[j]%mod*pr[j]+pr[j]+1)%mod;
    				sd2[i*pr[j]]=1ll*_sd2[i]*sum2[i*pr[j]]%mod;
    				_sd2[i*pr[j]]=_sd2[i];
    				mu[i*pr[j]]=0;
    				break;
    			}
    			sum[i*pr[j]]=1+pr[j];
    			sd[i*pr[j]]=1ll*sd[i]*sum[i*pr[j]]%mod;
    			_sd[i*pr[j]]=sd[i];
    			sum2[i*pr[j]]=(1ll*pr[j]*pr[j]+pr[j]+1)%mod;
    			sd2[i*pr[j]]=1ll*sd2[i]*sum2[i*pr[j]]%mod;
    			_sd2[i*pr[j]]=sd2[i];
    			mu[i*pr[j]]=mod-mu[i];
    		}
    	}
    	for(int i=1;i<=N;++i)ssd[i]=(ssd[i-1]+sd[i])%mod;
    	for(int i=1;i<=N;++i)ssd2[i]=(ssd2[i-1]+1ll*i*sd2[i])%mod;
    	for(int i=1;i<=N;++i)
    		for(int j=i;j<=N;j+=i)
    			if(mu[i])ans[j]=(ans[j]+2ll*j*mu[i]%mod*i%mod*sd[j/i]%mod*ssd[j/i])%mod;
    	for(int i=1;i<=N;++i)ans[i]=(ans[i]+ans[i-1])%mod;
    	int T=gi(),n;
    	for(int i=1;i<=T;++i)n=gi(),printf("Case #%d: %d
    ",i,(ans[n]-ssd2[n]+mod)%mod);
    	return 0;
    }
    
    

    右边用推式子:

    #include<bits/stdc++.h>
    #define il inline
    #define vd void
    #define mod 1000000007
    typedef long long ll;
    il ll gi(){
    	ll x=0,f=1;
    	char ch=getchar();
    	while(!isdigit(ch))f^=ch=='-',ch=getchar();
    	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    	return f?x:-x;
    }
    int pr[1000010],P,sd[1000010],_sd[1000010],sum[1000010],ssd[1000010],mu[1000010];
    bool yes[1000010];
    int ans[1000010];
    int main(){
    #ifdef XZZSB
    	freopen("in.in","r",stdin);
    	freopen("out.out","w",stdout);
    #endif
    	int N=1000000;
    	sd[1]=1;mu[1]=1;
    	for(int i=2;i<=N;++i){
    		if(!yes[i])pr[++P]=i,sd[i]=i+1,_sd[i]=1,sum[i]=i+1,mu[i]=mod-1;
    		for(int j=1;j<=P&&i*pr[j]<=N;++j){
    			yes[i*pr[j]]=1;
    			if(i%pr[j]==0){
    				sum[i*pr[j]]=(1ll*sum[i]*pr[j]+1)%mod;
    				sd[i*pr[j]]=1ll*_sd[i]*sum[i*pr[j]]%mod;
    				_sd[i*pr[j]]=_sd[i];
    				mu[i*pr[j]]=0;
    				break;
    			}
    			sum[i*pr[j]]=1+pr[j];
    			sd[i*pr[j]]=1ll*sd[i]*sum[i*pr[j]]%mod;
    			_sd[i*pr[j]]=sd[i];
    			mu[i*pr[j]]=mod-mu[i];
    		}
    	}
    	for(int i=1;i<=N;++i)ssd[i]=(ssd[i-1]+sd[i])%mod;
    	for(int i=1;i<=N;++i)
    		for(int j=i;j<=N;j+=i){
    			ans[j]=(ans[j]+2ll*j*mu[j/i]%mod*(j/i)%mod*sd[i]%mod*ssd[i])%mod;
    			ans[j]=(ans[j]-1ll*j*mu[i]%mod*i%mod*sd[j/i]%mod*sd[j/i]%mod+mod)%mod;
    		}
    	for(int i=1;i<=N;++i)ans[i]=(ans[i]+ans[i-1])%mod;
    	int T=gi(),n;
    	for(int i=1;i<=T;++i)n=gi(),printf("Case #%d: %d
    ",i,ans[n]);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/xzz_233/p/11142149.html
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