Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15023		Accepted: 5962

The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are 
F2 = {1/2} 
F3 = {1/3, 1/2, 2/3} 
F4 = {1/4, 1/3, 1/2, 2/3, 3/4} 
F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5}

You task is to calculate the number of terms in the Farey sequence Fn.

There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 10⁶). There are no blank lines between cases. A line with a single 0 terminates the input.

For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn. 

2

3

4

5

0

1

3

5

9

POJ Contest,Author:Mathematica@ZSU

 /*by SilverN*/

 #include<iostream>

 #include<algorithm>

 #include<cstring>

 #include<cstdio>

 #include<cmath>

 using namespace std;

 long long f[];

 int n;

 void phi(){

     int i,j;

     for(i=;i<=;i++)

         if(!f[i])

             for(j=i;j<=;j+=i){

                 if(!f[j])f[j]=j;

                 f[j]=f[j]/i*(i-);

             }

 }

 int main(){

     phi();

     for(int i=;i<=;i++){

         f[i]+=f[i-];//求前缀和

     }

     while(scanf("%d",&n) && n){

         cout<<f[n]<<endl;

     }

     return ;

 }