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  • 51nod 1238 最小公倍数之和 V3

    51nod 1238 最小公倍数之和 V3

    [sum_{i=1}^Nsum_{j=1}^N lcm(i,j) ]

    (Nleq 10^{10})

    先按照套路推一波反演的式子:

    [Ans=sum_{g=1}gsum_{i=1}^{frac{n}{g}}sum_{j=1}^{frac{n}{g}}ijsum_{d|i,d|j}mu(d)\ =sum_{g=1}gsum_{d=1}^{frac{n}{g}}d^2mu(d)S^2(frac{n}{dg})\ =sum_{T=1}^nsum_{d|T}d^2mu(d)frac{T}{d}S^2(frac{n}{T})\ ]

    难点在于求下面的函数的前缀和。

    [G(n)=sum_{d|T}d^2mu(d)frac{T}{d} ]

    设:

    [A(n)=n^2mu(n)\ B(n)=n ]

    则:

    [G(n)=A*B ]

    其中(*)表示狄利克雷卷积。

    考虑用杜教筛,也就是构造一个函数(C(n)),使得(G*C)有些美妙的性质。

    考虑从(A(n)=n^2mu(n))下手,将(n^2)消掉,只留下(mu(n))

    (C(n)=n^2),

    [A*C=sum_{d|n}d^2mu(d)(frac{n}{d})^2\ =n^2sum_{d|n}mu(d)\ =[n==1] ]

    所以:

    [egin{align} G*C&=(A*C)*B\ &=epsilon *B\ &=sum_{d=1}[d==1]frac{n}{d}\ &=n end{align} ]

    然后就是杜教筛的套路:

    [sum_{i=1}^nsum_{d|n}G(n)(frac{n}{d})^2=sum_{i=1}^ni\ Rightarrow sum_{i=1}^ni^2sum_{j=1}^{lfloorfrac{n}{i} floor}G(j)=frac{n*(n+1)}{2}\ Rightarrow sum_{i=1}^ni^2S_G(lfloorfrac{n}{i} floor)=frac{n*(n+1)}{2}\ Rightarrow S_G(n)=frac{n*(n+1)}{2}-sum_{i=2}^ni^2S_G(lfloorfrac{n}{i} floor)\ ]

    代码:

    #include<bits/stdc++.h>
    #define ll long long
    #define maxx 3000005
    
    using namespace std;
    inline ll Get() {ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}while('0'<=ch&&ch<='9') {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}return x*f;}
    
    const ll mod=1e9+7;
    
    ll ksm(ll t,ll x) {
    	ll ans=1;
    	for(;x;x>>=1,t=t*t%mod)
    		if(x&1) ans=ans*t%mod;
    	return ans;
    }
    
    ll n;
    int p[maxx];
    ll f[maxx];
    bool vis[maxx];
    const ll inv2=ksm(2,mod-2),inv6=ksm(6,mod-2);
    ll cal(ll n) {return n*(n+1)%mod*inv2%mod;}
    ll cal2(ll n) {return n*(n+1)%mod*(2*n+1)%mod*inv6%mod;}
    
    void pre(int n) {
    	for(int i=2;i<=n;i++) {
    		if(!vis[i]) p[++p[0]]=i,f[i]=1-i+mod;
    		for(int j=1;j<=p[0]&&1ll*i*p[j]<=n;j++) {
    			vis[i*p[j]]=1;
    			if(i%p[j]==0) {
    				f[i*p[j]]=f[i];
    				break;
    			}
    			f[i*p[j]]=(1-p[j])*f[i]%mod;
    		}
    	}
    	f[1]=1;
    	for(int i=1;i<=n;i++) {
    		f[i]=((f[i]*i+f[i-1])%mod+mod)%mod;
    	}
    }
    
    map<ll,ll>st;
    ll Sum(ll n) {
    	if(n<=3000000) return f[n];
    	if(st.find(n)!=st.end()) return st[n];
    	ll ans=cal(n%mod);
    	ll last=1;
    	for(ll i=2;i<=n;i=last+1) {
    		ll now=n/(n/i);
    		ans=(ans-(cal2(now%mod)-cal2(last%mod)+mod)*Sum(n/i)%mod+mod)%mod;
    		last=now;
    	}
    	return st[n]=ans;
    }
    
    ll solve(ll n) {
    	ll ans=0;
    	ll last=0;
    	for(ll i=1;i<=n;i=last+1) {
    		ll now=n/(n/i);
    		(ans+=(Sum(now)-Sum(last)+mod)*cal(n/i%mod)%mod*cal(n/i%mod))%=mod;
    		last=now;
    	}
    	return ans;
    }
    
    int main() {
    	pre(3000000);
    	n=Get();
    	cout<<solve(n);
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/hchhch233/p/10645777.html
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