luogu P2568 GCD

所以洛谷两道这样的题,另一道要用MI?

考虑每个质数对答案的贡献

如果有\(gcd(a,b)=p\),则显然\(gcd(\frac{a}{p},\frac{b}{p})=1\)

所以每个\(p\)对答案的贡献就是\(\sum_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{p}\rfloor}[gcd(i,j)=1]\)

若\(i>j\),对于每个\(i\),贡献为\((\sum_{j=1}^{i-1}[gcd(i,j)=1])=\phi[i]\)

对于\(i<j\)的情况也是一样的,也就是每个单独的\(j\)贡献为\(\phi[j]\)

若\(i=j\),则有\(1\)的贡献

最终答案即为\((\sum_{i=1}^{m}\sum_{j=1}^{\lfloor {n/p_i}\rfloor}\phi[j])+m(m\)为质数个数,\(p_i\)为第\(i\)个质数\()\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<cmath>
#include<ctime>
#include<queue>
#include<map>
#define LL long long
#define il inline
#define re register

using namespace std;
const LL mod=1000000007;
il LL rd()
{
    re LL x=0,w=1;re char ch;
    while(ch<'0'||ch>'9') {if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') {x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
    return x*w;
}
LL n,phi[10000010],ans;
int prm[1000010],tt;
char v[10000010];
il void init()
{
  phi [1] = 1; 
  for(re int i=2;i<=n;i++)
    {
      if(!v[i]) prm[++tt]=i,phi[i]=i-1;
      for(re int j=1;j<=tt&&i*prm[j]<=n;j++)
        {
          v[i*prm[j]]=1,phi[i*prm[j]]=phi[i]*(prm[j]-1);
          if(i%prm[j]==0) {phi[i*prm[j]]+=phi[i];break;}
        }
    }
}
il void work()
{
  for(re int i=1;i<=n;i++) phi [i] + = phi [i-1];  
  for(re int i=1;i<=tt;i++) ans+=phi[n/prm[i]];
  printf("%lld\n",(ans<<1)-tt); //知道为什么是-而不是+吗? 嘿嘿嘿
}


int main()
{
  n=rd();
  init();
  work();
  return 0;
}

luogu P2568 GCD

原文：https://www.cnblogs.com/smyjr/p/9409486.html