题意:求$\sum_{i=1}^n \sum_{j=1}^n gcd(i,j),n<=1e10$
之前刚好在UVA上也做过一个这样求和的题目,不过那个数据范围比较小,一开始用类似的方法
$ans=\sum_{i=1}^n \sum_{j=1}^i gcd(i,j)-\sum_{i=1}^n i$
先考虑化简$\sum_{i=1}^n gcd(i,n)$变成好求和的形式
$$\begin{aligned} \sum_{i=1}^n gcd(i,n) &=\sum_{i=1}^n \sum_{d=1}^n d*[gcd(i,n)=d]\\ &=\sum_{d=1}^n d \sum_{i=1}^n [gcd(i,n)=d] \\ &=\sum_{d=1}^n d \sum_{\frac{i}{d}=1}^{\frac{n}{d}} [gcd(\frac{i}{d},\frac{n}{d})=1]\\ &=\sum_{d|n} d*\phi(\frac{n}{d}) \end{aligned}$$
发现是$f(n)=n$和$g(n)=\phi(n)$的卷积,令$S(n)$表示$g$的前缀和,然后非常套路地,然后刚刚那一坨的前缀和就变成求$\sum_{i=1}^n i*S(\lfloor \frac{n}{i} \rfloor)$,欧拉函数前缀和直接用杜教筛算,然后分块求和
一开始取模一直写挂…orz
(有点懒直接用map存了)
#include<cstdio> #include<cstring> #include<map> using namespace std; typedef long long lint; const lint MOD=1000000007; const lint N=4000005; const lint G=100005; lint n,tot,inv2; lint pri[N],phi[N]; bool p[N]; map<lint,lint>mp; inline void init() { p[1]=1;phi[1]=1; for(register lint i=2;i<N;i++) { if(!p[i]) { pri[++tot]=i; phi[i]=i-1; } for(register lint j=1;j<=tot&&i*pri[j]<N;j++) { lint t=i*pri[j];p[t]=1; if(i%pri[j]==0){phi[t]=phi[i]*pri[j];break;} phi[t]=phi[i]*(pri[j]-1); } } for(register lint i=1;i<N;i++)phi[i]=(phi[i]+phi[i-1])%MOD; } inline lint pow_mod(lint a,lint b,lint p) { lint res=1; for(;b;b>>=1,a=(a*a)%p)if(b&1)res=(res*a)%p; return res%p; } inline lint sum(lint x) { return x%MOD*((x+1)%MOD)%MOD*inv2%MOD; } inline lint calc_phi(lint x) { if(x<N)return phi[x]; if(mp.count(x))return mp[x]; lint res=sum(x),pos; for(register lint i=2;i<=x;i=pos+1) { pos=x/(x/i); res-=((pos-i+1)%MOD*calc_phi(x/i)%MOD)%MOD; res=(res%MOD+MOD)%MOD; }return mp[x]=res; } inline lint calc_ans(lint x) { lint res=0,pos; for(register lint i=1;i<=x;i=pos+1) { pos=x/(x/i); res+=(sum(pos)-sum(i-1))%MOD*calc_phi(x/i)%MOD; res=(res%MOD+MOD)%MOD; }return res; } int main() { init();inv2=pow_mod(2,MOD-2,MOD); scanf("%lld",&n);lint ans=calc_ans(n)%MOD; ans=(ans*2)%MOD-sum(n);ans=(ans%MOD+MOD)%MOD; printf("%lld",ans); return 0; }