BZOJ 2671 Calc

题目链接

https://lydsy.com/JudgeOnline/problem.php?id=2671

题解

对于一对$(a,b)$满足$a+b\mid ab$，假设$d=\gcd(a,b),a=xd,b=yd$，那么
$xd+yd\mid xyd^2\\ x+y\mid xyd\\ x+y\mid d$
设$m=\lfloor\sqrt{n}\rfloor$答案就是
$\begin{aligned} & \sum_{j=1}^{m}\sum_{i=1}^{j-1}\lfloor\frac{n}{i+j}\rfloor [\gcd(i,j)=1]\\ = & \sum_{d=1}^{m}\mu(d)\sum_{j=1}^{\lfloor m/d\rfloor}\sum_{i=1}^{j-1}\lfloor\frac{n}{d^2(i+j)}\rfloor \end{aligned}$
复杂度大概是$O(n^{3/4})$的，这个请大家自行证明。

代码

#include <cmath>
#include <cstdio>
#include <algorithm>

int read()
{
  int x=0,f=1;
  char ch=getchar();
  while((ch<'0')||(ch>'9'))
    {
      if(ch=='-')
        {
          f=-f;
        }
      ch=getchar();
    }
  while((ch>='0')&&(ch<='9'))
    {
      x=x*10+ch-'0';
      ch=getchar();
    }
  return x*f;
}

const int maxn=65536;

int p[maxn+10],prime[maxn+10],cnt,mu[maxn+10];

int getprime()
{
  p[1]=mu[1]=1;
  for(int i=2; i<=maxn; ++i)
    {
      if(!p[i])
        {
          prime[++cnt]=i;
          mu[i]=-1;
        }
      for(int j=1; (j<=cnt)&&(i*prime[j]<=maxn); ++j)
        {
          int x=i*prime[j];
          p[x]=1;
          if(i%prime[j]==0)
            {
              mu[x]=0;
              break;
            }
          mu[x]=-mu[i];
        }
    }
  return 0;
}

long long calc(int n,int d)
{
  long long ans=0;
  for(int i=1; i<=d; ++i)
    {
      int t=n/i;
      for(int l=i+1,r; (l<(i<<1))&&(l<=t); l=r+1)
        {
          r=std::min((i<<1)-1,t/(t/l));
          ans+=1ll*(r-l+1)*(t/l);
        }
    }
  return ans;
}

int n,s;

int main()
{
  getprime();
  n=read();
  s=sqrt(n);
  long long ans=0;
  for(int i=1; i<=s; ++i)
    {
      ans+=mu[i]*calc(n/i/i,s/i);
    }
  printf("%lld\n",ans);
  return 0;
}