51nod 1594 Gcd and Phi 反演

OTZ 又被吊打了。。。我当初学的都去哪了？？？

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思路：反演套路？

提交：\(1\)次

题解：

求\(\sum_{i=1}^{n}\sum_{j=1}^{n}\varphi(gcd(\varphi(i),\varphi(j)))\)

设\(c[i]=\sum_{j=1}^n[\varphi(j)==i]\)

有：

\(\sum_{i=1}^{n}\sum_{j=1}^{n}\varphi(gcd(i,j))*c[i]*c[j]\)

欧拉反演一下，把\(gcd\)扔出来

\(\sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{d|gcd(i,j)}\varphi(d)*c[i]*c[j]*[gcd(i,j)==d]\)

$\sum_{d=1}^{n}\varphi(d) \sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\ \sum_{j=1}^{\lfloor \frac{n}{d} \rfloor} c[id] * c[jd] * [gcd(i,j)==1] $

$\sum_{d=1}^{n}\varphi(d) \sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}c[id] \sum_{j=1}^{\lfloor \frac{n}{d} \rfloor}c[jd] \ [gcd(i,j)==1] $

\(\sum_{d=1}^{n}\varphi(d) \sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}c[i*d] \sum_{j=1}^{\lfloor \frac{n}{d} \rfloor}c[j*d] \sum_{k|gcd(i,j)} \mu(k)\)

\(\sum_{d=1}^{n}\varphi(d) \sum_{k=1}^{\lfloor \frac{n}{d} \rfloor} \mu(k) \sum_{i=1}^{\lfloor \frac{n}{d*k} \rfloor}c[i*d*k] \sum_{j=1}^{\lfloor \frac{n}{d*k} \rfloor}c[j*d*k]\)

设\(s[i]=\sum_{j=1}^{\lfloor \frac{n}{i} \rfloor}\ c[i*j]\),注意到\(s[i]\)可以在\(O(nlogn)\)计算出来

\(\sum_{d=1}^{n}\ \varphi(d) \sum_{k=1}^{\lfloor \frac{n}{d} \rfloor} \mu(k)\ s[d*k]^2\)

交换一下\(d\)与\(k\)，可以优化枚举（见代码）

\(\sum_{d=1}^{n}\ \mu(d) \sum_{k=1}^{\lfloor \frac{n}{d} \rfloor} \varphi(k)\ s[d*k]^2\)

#include<cstdio>
#include<iostream>
#include<cstring>
#define ll long long
#define R register ll
using namespace std;
namespace Luitaryi {
template<class I> inline I g(I& x) { x=0;
  register I f=1; register char ch; while(!isdigit(ch=getchar())) f=ch=='-'?-1:f;
  do x=x*10+(ch^48); while(isdigit(ch=getchar())); return x*=f;
} const int N=2e6;
int T,n,cnt,p[N>>1],phi[N+10],mu[N+10],c[N+10];
ll s[N+10],ans; bool v[N+10];
inline void PRE() { phi[1]=mu[1]=1;
  for(R i=2;i<=N;++i) {
    if(!v[i]) p[++cnt]=i,phi[i]=i-1,mu[i]=-1;
    for(R j=1;j<=cnt&&i*p[j]<=N;++j) {
      v[i*p[j]]=true;
      if(i%p[j]==0) {
        phi[i*p[j]]=phi[i]*p[j];
        break;
      } phi[i*p[j]]=phi[i]*(p[j]-1);
      mu[i*p[j]]=-mu[i];
    }
  }
}
inline void calc(int n) { ans=0;
  memset(c,0,sizeof(c)),memset(s,0,sizeof(s));
  for(R i=1;i<=n;++i) ++c[phi[i]];
  for(R i=1;i<=n;++i) for(R j=1,lim=n/i;j<=lim;++j) s[i]+=c[i*j];
  for(R d=1;d<=n;++d) if(mu[d]) for(R k=1,lim=n/d;k<=lim;++k) //if(mu[d]) 优化枚举
    ans+=mu[d]*phi[k]*s[d*k]*s[d*k];
}
inline void main() {g(T); PRE(); while(T--) g(n),calc(n),printf("%lld\n",ans);}
} signed main() {Luitaryi::main(); return 0;}

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2019.08.11

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