D

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.

Input

There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 10 6). There are no blank lines between cases. A line with a single 0 terminates the input.

Output

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

Sample Input

2
3
4
5
0

Sample Output

1
3
5
9

欧拉函数打表模板

复杂度O（n*loglogn）

#include<stdio.h>
using namespace std;
long long phi[1000005];
void getphi(long long maxn){
    for(int i = 2; i < maxn; i++){
        phi[i] = i;
    }
    phi[1] = 1;
    for(int i = 2; i < maxn; i++){
        if(phi[i] == i){
            for(long long j = i; j < maxn; j += i){
                phi[j] = phi[j] / i * (i - 1);
            }
        }
    }
}
long long sum[1000005];
int main()
{
    sum[0]=0;
    getphi(1000003);
    for(long long i=1;i<=1000000;i++)
    {
        sum[i]=sum[i-1]+phi[i];
    }
    int n;
    while(~scanf("%lld",&n))
    {
        if(n==0) break;
        printf("%lld
",sum[n]-1);
    }
}