正题

题目链接:http://www.51nod.com/Challenge/Problem.html#problemId=1227

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题目大意

定义
$$F(a)=\frac{{\sum }_{i=1}^{a}lcm(a,i)}{a}$$
给出$l,r$求${\sum }_{i=l}^{r}F(i)$

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解题思路

好久没做数论题了
直接拆成两个前缀和的差，然后有
$$\sum _{i=1}^n\sum _{j=1}^i\frac{j}{gcd(i,j)}$$
$$\sum _{d=1}^n\frac{1}{d}\sum _{i=1}^n\sum _{j=1}^{i}j[gcd(i,j)=d]$$
$$\sum _{d=1}^n\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{i}j[gcd(i,j)=1]$$
$$\frac{1}{2}+\sum _{d=1}^n\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\frac{\phi (i)i}{2}$$

函数$H(n)=\phi (n)n$可以用杜教筛，$H\times id$就可以得到$n^2$的函数。

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code

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#define ll long long
using namespace std;
const ll N=5e6,P=1e9+7;
ll cnt,pri[N/10],phi[N];
bool v[N];map<ll,ll> mp;
void Prime(){phi[1]=1;for(ll i=2;i<N;i++){if(!v[i])pri[++cnt]=i,phi[i]=i-1;for(ll j=1;j<=cnt&&i*pri[j]<N;j++){v[i*pri[j]]=1;if(i%pri[j]==0){phi[i*pri[j]]=phi[i]*pri[j];break;}phi[i*pri[j]]=phi[i]*(pri[j]-1);}}for(ll i=1;i<N;i++)phi[i]=(phi[i]*i+phi[i-1])%P;return;
}
ll GetS(ll n)
{return n*(n+1)/2%P;}
ll GetSS(ll n)
{return n*(n+1)%P*(2*n+1)%P*((P+1)/6)%P;}
ll GetPhi(ll n){if(n<N)return phi[n];if(mp[n])return mp[n];ll ans=GetSS(n);for(ll l=2,r;l<=n;l=r+1){r=n/(n/l);(ans-=GetPhi(n/l)*(GetS(r)-GetS(l-1))%P)%=P;}mp[n]=ans;return ans;
}
ll solve(ll n){ll ans=0;if(!n)return 0;for(ll l=1,r;l<=n;l=r+1){r=n/(n/l);(ans+=(GetPhi(n/l)+1)*(r-l+1)%P)%=P;}return ans*((P+1)/2)%P;
}
signed main()
{ll l,r,ans;Prime();scanf("%lld%lld",&l,&r);ans=(solve(r)-solve(l-1))%P;printf("%lld\n",(ans+P)%P);return 0;
}