"BZOJ2190" "SDOI2008" honor Guard

Description

As a sports Commissioner, C June is responsible for the training of the honor guard. 　　  　　Guard of Honor is composed of students of N * N of the square, in order to ensure that the procession uniform, c June will follow the guard of Honor's left rear, according to the number of students in the line of sight to determine whether the team is neat (such as). Now, C-June wants you to tell him the number of students you can see when the team is neat.

Input

A total of a number n.

Output

A total of a number, that is, c June should see the number of students.

Sample Input4

Sample Output9


HINT

"Data size and conventions" for 100% of data, 1≤n≤40000

Source

Arithmetical

idea: Look at the picture, is symmetrical, with the diagonal as the axis of symmetry, not to see the special three points, the calculation found that the points on the line to meet the Euler function, so the linear sieve to 2~n-1 Eulerian function can, finally do not forget the symmetry and special three points. Time 10sec not linear sieve also OK?

1#include <iostream>2#include <cstdio>3#include <cstring>4 #defineN 400005 using namespacestd;6 intflag[n+Ten],prime[n+Ten],phi[n+Ten];7 intN,ans;8 voidErphi ()9 {Ten     intk=1; Onememset (Flag,0,sizeof(flag)); A      for(intI=2; i<=n;i++) -     { -         if(!flag[i]) prime[k++]=i,phi[i]=i-1; the          for(intj=1; j<k&&i*prime[j]<n;j++) -         { -flag[i*prime[j]]=true; -             if(i%prime[j]==0)     +             { -phi[i*prime[j]]=phi[i]*Prime[j]; +                  Break; A             } at             Elsephi[i*prime[j]]=phi[i]* (prime[j]-1); -         } -     } -     return; - } - intMain () in { -scanf"%d",&n); to Erphi (); +      for(intI=2; i<n;i++) -ans+=Phi[i]; theprintf"%d\n", ans*2+3); *     return 0; $}

"BZOJ2190" "SDOI2008" honor Guard