Bzoj 2818 GCD

The main point of this paper is to use linear sieve method to find Euler function φ, first of all, say what is Euler function: The positive integer n, the Euler function is less than or equal to n the number of the number of n coprime. The reason to use the linear sieve is because it allows us to find the value of O (n).

First, a brief introduction to the linear sieve method to find Euler function φ: (Excerpt from Baidu)

intm[maxn],phi[maxn],p[maxn],pt;//M[i] is the minimum factor of I, P is prime, PT is a number of prime intMake () {intk,n=MAXN; phi[1]=1;  for(intI=2; i<n;i++)    {        if(!m[i])//I is the prime number{p[pt++]=m[i]=i; Phi[i]=i-1; }         for(intj=0;j<pt&& (k=p[j]*i) <n;j++) {M[k]=P[j]; if(M[i]==p[j])//in order to ensure that the future number is not re-screened, to break{Phi[k]=phi[i]*P[j];/*here Phi[k] and Phi[i] behind the ∏ (p[i]-1)/p[i] All the same (M[i]==p[j]) only one p[j], you can guarantee ∏ (p[i]-1)/p[i] in front of the same*/                 Break; }            Else{Phi[k]=phi[i]* (p[j]-1);//properties of the integrable function, F (i*k) =f (i) *f (k)            }        }    }}

After that, I'm going to go long. Find out the prefix of φ and multiply by 2 minus 1, it is good, the reduction is more than the calculation.

The code for this problem is as follows:

#include <cstdio>#include<cstdlib>#include<iostream>#include<cstring>#include<algorithm>#include<cmath>using namespacestd; typedefLong Longll;Constll maxn=10000005; ll I_number,m;ll F[MAXN],B[MAXN],P[MAXN]; intMain () {CIN>>I_number; f[1]=1;  for(LL i=2; i<=i_number;i++)    {        if(!B[i]) {P[m++]=i; F[i]=i-1; }                  for(LL j=0; j<m&&p[j]*i<=i_number;j++) {F[p[j]*i]=f[i]*P[j]; B[P[J]*i]=1; if(! (I%p[j])) Break; F[P[J]*i]= (p[j]-1)*F[i]; }} ll I_temp=0;  for(LL i=1; i<=i_number;i++) {F[i]=f[i]+f[i-1]; }       for(LL i=1; i<=i_number;i++) {F[i]=f[i]*2-1; }       for(LL i=0; i<m;i++) {i_temp=i_temp+f[i_number/P[i]]; } cout<<i_temp<<Endl; return 0;}

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Bzoj 2818 GCD