Sdoi2008 Guard of Honor

Title Description 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.
　　

Enter a description input Description
A total of a number n.

outputs description output Description
A total of a number, that is, c June should see the number of students.

sample input to sample
4

sample output Sample outputs
9

Data Size & Hint
For 30% of data, 1≤n≤1000
For 100% of data, 1≤n≤40000

Solving
The lower left corner is (0,0), if (x, y) is visible, then gcd (x, y) = 1, analytic geometry or two-dimensional vector something to think about, you can figure it out.
Press Y=x to divide the coordinate system into two halves. For each half, the X column always has phi (x) visible, and (similarly, the Y line always has phi (Y) to be seen.) These two are available with one) (Phi for Euler functions)
This will add Phi (2~n-1), multiply by 2, plus (0,1) (1,0) (three) points can be answered

• Code:

#include <cstdio>#include <algorithm>#define LL unsigned long longusing namespace STD;Const intMAXN =40050;intN;ll Phi[maxn], ans;voidReadint&a) {intf =1;Charch = getchar (); A =0; while(Ch <' 0 '|| CH >' 9 ')    {if(ch = ='-') F =-1;    ch = getchar (); } while(Ch >=' 0 '&& CH <=' 9 ') {a = a *Ten+ CH- -;    ch = getchar (); } a *= F;}voidWrite (ll a) {inttop =0;Charch[ -];if(A <0)    {Putchar('-');    A =-A; } Do{ch[top++] = a%Ten+ -; A/=Ten; } while(a); while(top--)Putchar(Ch[top]);Putchar(' \ n ');}voidEulerphi (intA//Sieve method to find Euler function{phi[1] =1; for(inti =2; I <= A; ++i)if(Phi[i] = =0) for(intj = i; J <= A; J + = i) {if(Phi[j] = =0) Phi[j] = j; PHI[J] = phi[j]/I * (i-1); }}voidInit () {read (n); Eulerphi (n);}voidWork () { for(inti =2; I < n;    ++i) ans + = phi[i]; Write (ANS *2+3);}intMain () {init (); Work ();return 0;}

Sdoi2008 Guard of Honor