【BZOJ 2671】Calc

题目要求求出$\sum_{a=1}^{N}\sum_{b=a+1}^{N}[a+b|ab]$
我们设$d=gcd(a,b),a=nd,b=md$则转化为
$[nd+md|nmd^2]=[n+m|nmd]$又有$n,m$互质，则显然有$n+m|d$
原式可推为
$\sum_{d=1}^{N}\sum_{n=1}^{\lfloor {N \over d}\rfloor}\sum_{m=n+1}^{\lfloor {N \over d}\rfloor}e(gcd(n,m))[n+m|d]$
[n+m|d]的个数显然为$\lfloor {N \over m(n+m)} \rfloor$
$e(gcd(n,m))$显然可以莫比乌斯反演
所以进一步推为
$=\sum_{m=1}^{\sqrt N }\sum_{n=1}^{m-1}\lfloor {N \over m(n+m)} \rfloor \sum_{d|n,d|m}\mu(d)$
用$md,nd$来替代$m,n$则有
$\sum_{d=1}^{\sqrt N} \mu(d)\sum_{m=1}^{\lfloor {\sqrt N \over d}\rfloor}\sum_{n=1}^{m-1} \lfloor{\lfloor {N \over d^2m}\rfloor\over n+m} \rfloor$
于是可以分块求得，code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
long long n,sqn,tot;
long long mu[100001],prime[10001];
bool a[100001];
void prework(long long n)
{
    long long i,j;
    mu[1]=1;
    for (i=2;i<=n;++i)
      {
        if (!a[i])
          {mu[i]=-1; prime[++tot]=i;}
        for (j=1;j<=tot&&i*prime[j]<=n;++j)
          {
            a[i*prime[j]]=true;
            if (i%prime[j]==0)
              {
                mu[i*prime[j]]=0;
                break;
              }
            mu[i*prime[j]]=-mu[i];
          }
      }
}
long long work(long long fz,long long n)
{
    long long ans=0,i,last;
    if (n==0) return 0;
    for (i=n+2;i<=2*n+1;i=last+1)
      {
        if (fz/i==0) break;
        last=min(2*n+1,fz/(fz/i));
        ans=ans+(last-i+1)*(fz/i);
      }
    return ans;
}
int main()
{
    long long d,m,ans=0,t;
    scanf("%lld",&n);
    sqn=sqrt(n);
    prework(sqn);
    for (d=1;d<=sqn;++d)
      {
        if (mu[d]==0) continue;
        t=0;
        for (m=2;m<=sqn/d;++m)
          t+=work(n/(d*d*m),m-1);
        ans=ans+mu[d]*t;
      }
    printf("%lld\n",ans);
}