莫比乌斯反演复健：Problem b（由于太颓废了也是6.30-7.6的周报）

莫比乌斯反演复健：Problem b（由于太颓废了也是6.30-7.6的周报）

先来一个莫比乌斯反演常见的模板

\[\begin{aligned} &\sum_i^n\sum_j^m[\gcd(i,j)=1]\\ \iff&\sum_i^n\sum_j^m\sum_{d|\gcd(i,j)}\mu(d)\\ \iff&\sum_{d=1}^n\mu(d)\sum_i^n\sum_j^m[d|\gcd(i,j)]\\ \iff&\sum_{d=1}^n\mu(d)\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor\\ \end{aligned} \]

对于该问题，经过简单的充斥，等同于去求

\[\begin{aligned} &\sum_i^n\sum_j^m[\gcd(i,j)=k]\\ \iff&\sum_i^{\lfloor\frac{n}{k}\rfloor}\sum_j^{\lfloor\frac{m}{k}\rfloor}[\gcd(i,j)=1]\\ \iff&\sum_i^{\lfloor\frac{n}{k}\rfloor}\sum_j^{\lfloor\frac{m}{k}\rfloor}\sum_{d|\gcd(i,j)}\mu(d)\\ \iff&\sum_{d=1}^n\mu(d)\sum_i^{\frac{n}{k}} \sum_j^{\frac{m}{k}}[d|\gcd(i,j)]\\ \iff&\sum_{d=1}^n\mu(d)\lfloor\frac{n}{kd} \rfloor\lfloor\frac{m}{kd}\rfloor\\ \end{aligned} \]

用整数分块和容斥原理处理即可

f(a,b,c,d)=g(b,d)-g(a,c-1)-g(a-1,c)+g(a-1,c-1)

代码如下：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define MAXN 50005
LL n,p,cnt,sum[MAXN];
int mu[MAXN],prime[MAXN];
bool vis[MAXN];
void oula(LL n)
{
  mu[1]=1;
  for(int i=2;i<=n;i++)
  {
    if(!vis[i])
    {
      prime[++cnt]=i;
      mu[i]=-1;
    }
    for(int j=1;j<=cnt&&i*prime[j]<=n;j++)
    {
      vis[i*prime[j]]=1;
      if(i%prime[j])
      {
        mu[i*prime[j]]=-mu[i];
      }else{
        break;
      }
    }
  }
  for(int i=1;i<=n;i++)sum[i]=sum[i-1]+mu[i];
}
LL solve(LL n,LL m)
{
  if(n>m)swap(n,m);
  LL ret=0;
  for(LL l=1,r=0;l<=n;l=r+1)
  {
    r=min(n/(n/l),m/(m/l));
    ret+=(sum[r]-sum[l-1])*(n/l)*(m/l);
  }
  return ret;
}
int main()
{
  int T;
  scanf("%d",&T);
  oula(50000);
  while(T --> 0)
  {
    LL a,b,c,d,k;
    cin>>a>>b>>c>>d>>k;
    cout<<solve((a-1)/k,(c-1)/k)+solve(b/k,d/k)-solve((a-1)/k,d/k)-solve((c-1)/k,b/k)<<endl;
  }
  return 0;
}