BZOJ 2005 [Noi2010]能量采集

版权声明：本文为博主原创文章，未经博主允许可以转载，但要注明出处 https://blog.csdn.net/wang3312362136/article/details/85639056

题目链接

https://lydsy.com/JudgeOnline/problem.php?id=2005

题解

容易发现，植物 $(x,y)$和机器之间的连线上植物的数量就是 $\gcd(x,y)$。

因此答案就是
$\begin{aligned} & \sum_{i=1}^n \sum_{j=1}^m (2\gcd(i,j)-1)\\ = & 2\sum_{i=1}^n \sum_{j=1}^m \gcd(i,j)-nm \end{aligned}$
设
$\begin{aligned} g = & \sum_{i=1}^n\sum_{j=1}^m \gcd(i,j)\\ = & \sum_{T=1}^{\min(n,m)}\lfloor \frac{n}{T} \rfloor \lfloor\frac{m}{T}\rfloor \sum_{d|T}d\mu(\frac{d}{T})\\ = & \sum_{T=1}^{\min(n,m)}\lfloor\frac{n}{T}\rfloor \lfloor\frac{m}{T}\rfloor\varphi(T) \end{aligned}$
这个显然可以整除分块求，当然也可以直接枚举。

则答案就是
$\begin{aligned} & 2g+nm\\ = & 2\sum_{T=1}^{\min(n,m)}\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor\varphi(T)+nm \end{aligned}$

代码

#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=100000;

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

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

int n,m;
long long ans;

int main()
{
  getprime();
  n=read();
  m=read();
  for(int i=1; i<=std::min(n,m); ++i)
    {
      ans+=1ll*(n/i)*(m/i)*phi[i];
    }
  printf("%lld\n",2*ans-1ll*n*m);
  return 0;
}