所以洛谷兩道這樣的題,另一道要用MI?

考慮每個質數對答案的貢獻

如果有\(gcd(a,b)=p\),則顯然\(gcd(\frac{a}{p},\frac{b}{p})=1\)

所以每個\(p\)對答案的貢獻就是\(\sum_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{p}\rfloor}[gcd(i,j)=1]\)

若\(i>j\),對於每個\(i\),貢獻為\((\sum_{j=1}^{i-1}[gcd(i,j)=1])=\phi[i]\)

對於\(i<j\)的情況也是一樣的,也就是每個單獨的\(j\)

若\(i=j\),則有\(1\)的貢獻

最終答案即為\((\sum_{i=1}^{m}\sum_{j=1}^{\lfloor {n/p_i}\rfloor}\phi[j])+m(m\)為質數個數,\(p_i\)為第\(i\)個質數\()\)

#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); //知道為什麽是-而不是+嗎? 嘿嘿嘿
}


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

luogu P2568 GCD