51nod1220 约数之和

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1220

$G(n)=sumlimits_{i=1}^nsumlimits_{j=1}^nsumlimits_{k=1}^{n^2}[k|ij]cdot k$
$ small{[k|ij] o[frac{k}{gcd(i,k)}|j]}$
$G(n)=sumlimits_{i=1}^nsumlimits_{j=1}^nsumlimits_{k=1}^{n^2}[frac{k}{gcd(i,k)}|j]cdot k$
$ small{gcd(i,k) o g,i/g o i,sumlimits_{j=1}^X[Y|j] o lfloorfrac XY floor}$
$G(n)=sumlimits_{k=1}^{n^2}sumlimits_{g|k}sumlimits_{i=1}^{n/g}[gcd(i,k/g)=1]lfloorfrac{n}{k/g} floorcdot k$
$ small{k/g o k}$
$G(n)=sumlimits_{g=1}^{n}sumlimits_{k=1}^{n}sumlimits_{i=1}^{n/g}[gcd(i,k)=1]lfloorfrac{n}{k} floorcdot kcdot g$
$small{[X=1] osumlimits_{d|X}mu(d)}$
$G(n)=sumlimits_{g=1}^{n}sumlimits_{k=1}^{n}sumlimits_{i=1}^{n/g}sumlimits_{d|iwedge d|k}mu(d)lfloorfrac{n}{k} floorcdot kcdot g$
$small{i/d o i,k/d o k}$
$G(n)=sumlimits_{d=1}^nsumlimits_{g=1}^{n/d}sumlimits_{k=1}^{n/d}sumlimits_{i=1}^{n/gd}mu(d)lfloorfrac{n}{kd} floorcdot kcdot gcdot d$
$small{sumlimits_{i=1}^X1 o X}$
$G(n)=sumlimits_{d=1}^nsumlimits_{g=1}^{n/d}sumlimits_{k=1}^{n/d}mu(d)lfloorfrac{n}{gd} floorlfloorfrac{n}{kd} floorcdot kcdot gcdot d$
$small{ig(sumlimits_{i=1}^{n}ilfloorfrac{n}{d} floorig)^2 o F(n)}$
$G(n)=sumlimits_{d=1}^nmu(d)dcdot F(lfloorfrac{n}{d} floor)$
$small{G o sum g,F o sum f}$
$g(n)=sum limits_{d|n}mu(d)dcdot f(n/d)$
$f(n)=sum limits_{d|n}dcdot g(n/d)$
$small{sum f o F,sum g o G}$
$F(n)=sumlimits_{d=1}^ndcdot G(lfloorfrac{n}{d} floor)$
$G(n)=F(n)-sumlimits_{d=2}^ndcdot G(lfloorfrac{n}{d} floor)$

记录一下推公式过程。

最后用记忆化搜索就可以$O(n^{3/4})$过了，如果加上线性预处理可以做到$O(n^{2/3})$

#include<bits/stdc++.h>
typedef unsigned long long i64;
const int P=1e9+7;
int n,vG[32007],vg[32007],B;
int G(int n){
    int&ans=n<=B?vg[n]:vG[::n/n];
    if(ans)return ans;
    i64 s=0,sx=n*2;
    for(int l=1,r,c;l<n;l=r){
        r=n/(c=n/(l+1));
        i64 t=i64(r+l+1)*(r-l);
        sx+=t*c;
        s+=t%P*G(c);
        if(s>i64(1.5e19))s%=P;
    }
    sx=sx/2%P;
    s=(sx*sx%P-s%P*((P+1)/2)%P+P)%P;
    return ans=s;
}
int main(){
    scanf("%d",&n);
    B=sqrt(n);
    printf("%d
",G(n));
    return 0;
}