"Hebei Province team Mutual test" GCD Bzoj 2818

Description

Given an integer n, 1<=x,y<=n and gcd (x, y) are prime
Number pairs (x, y) how many pairs.

Input

An integer n

Output

Title

Sample Input4

Sample Output4HINT

Hint

For examples (2,2), (2,4), (3,3), (4,2)

1<=n<=10^7

Ideas

I've seen a lot of questions about GCD and mods lately.

Learned a lot from recent topics. =

First there is such a property: if A∈[1,n],b∈[1,m], then GCD (A, B) |k has (n/k) * (m/k) group.

So f[x] is gcd (A, B) ==k the number of groups, f[k]= (n/k) * (m/k)-f[2k]-f[3k]-f[4k] ...

For this question: It seems that it is not possible to live.

Then there is a different nature:

If the number of A,B∈[1,N],GCD (A, B) ==k is equivalent to the number of groups of A,B∈[1,N/K],GCD (A, b) ==1.

This is good to ask, is the 1->n of Phi value (Euler function) *2-1.

First, the number of each group must be counted two times, such as (3,5) and (5,3), so to * *. Then do not count two times, so again-1.

Then it is how to find all the prime numbers and Euler functions in 1-n. On-site learning linear sieve.

In fact, I totally do not understand AH. First of all, remember. The core code is as follows:

1phi[1]=1; memset (Is_prime,true,sizeof(Is_prime));2  for(intI=2; i<=n;i++){3     if(Is_prime[i]) {4phi[i]=i-1;5prime[++cnt]=i; 6     }    7      for(intj=1; j<=cnt&&i*prime[j]<=n;j++){8is_prime[i*prime[j]]=false;9        if(i%prime[j]!=0) phi[i*prime[j]]=phi[i]* (prime[j]-1);Ten        Else{ Onephi[i*prime[j]]=phi[i]*Prime[j]; A           Break; -        } -     } the}

Linear Sieve

It should have been written wrong. In order to strengthen the memory of dictation. If there is a problem, look at the following version of it, that is AC.

1#include <iostream>2#include <cstring>3#include <string>4#include <cstdio>5#include <cstdlib>6#include <cmath>7#include <algorithm>8#include <queue>9#include <stack>Ten#include <map> One#include <Set> A#include <list> -#include <vector> -#include <ctime> the#include <functional> - #definePRITNF printf - #defineScafn scanf - #defineSACNF scanf + #definefor (i,j,k) for (int i= (j); i<= (k);(i) + +) - #defineClear (a) memset (A,0,sizeof (a)) + using namespacestd; Atypedef unsignedintUint; at Const intinf=0x3fffffff; - ///==============struct declaration============== -  - ///==============var declaration================= - Const intmaxn=10000050; - intn,tot=0, ans=0; in intPRIME[MAXN]; - Long LongPHI[MAXN]; to BOOLIS_PRIME[MAXN]; + ///==============function declaration============ - voidInit (); the ///==============main code======================= * intMain () $ {Panax Notoginseng //#define FILE__ - #ifdef file__ theFreopen ("input","R", stdin); +Freopen ("Output","W", stdout); A #endif thescanf"%d",&n); + Init (); -     for(intI=1; i<=n;i++) phi[i]+=phi[i-1]; $    Long Longans=0; $     for(intI=1; i<=tot;i++) -ans+=phi[n/prime[i]]*2-1; -printf"%lld\n", ans); the    return 0; - }Wuyi ///================fuction code==================== the voidInit () { -memset (Is_prime,true,sizeof(Is_prime)); phi[1]=1; Wu     for(intI=2; i<=n;i++){ -       if(Is_prime[i]) { Aboutphi[i]=i-1; $prime[++tot]=i; -       } -        for(intj=1; j<=tot;j++){ -          if(i*prime[j]>n) Break; Ais_prime[i*prime[j]]=false; +          if(i%prime[j]==0){ thephi[i*prime[j]]=phi[i]*Prime[j]; -              Break; $          } the          Else thephi[i*prime[j]]=phi[i]* (prime[j]-1); the       } the    } -}

BZOJ2818

Don't ask me why those qualities are. I also Phuket (╯‵-′) ╯︵┻━┻

"Hebei Province team Mutual test" GCD Bzoj 2818