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

思路：由a / b,gcd(a,b)=1知，当前f(n)是在f（n - 1）的基础上加上以n为分母，与n互质的数为分子的分数，所以f（n）比f（n - 1）增加了1~n内与n互质的数的个数，现在题意就很明显了，就是要求欧拉函数，不过需要一个前缀和。
代码实现如下：

 #include <cstdio>

 #include <cstring>

 #include <iostream>

 using namespace std;

 #define debug(x) cout <<'[' <<x <<']' <<endl;

 const int maxn = 1e6 + ;

 int n, m;

 int v[maxn], p[maxn], phi[maxn];

 long long sum[maxn];

 void euler() {

     m = ;

     memset(v, , sizeof(v));

     memset(sum, , sizeof(sum));

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

         if(v[i] == ) {

             v[i] = i;

             p[m++] = i;

             phi[i] = i - ;

         }

         for(int j = ; j < m; j++) {

             if(p[j] > v[i] || p[j] > maxn / i) break;

             v[i * p[j]] = p[j];

             phi[i * p[j]] = phi[i] * (i % p[j] ? p[j] -  : p[j]);

         }

     }

     phi[] = phi[] = ;

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

         sum[i] = sum[i - ] + phi[i];

     }

 }

 int main() {

     euler();

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

 //        debug(phi[n]);

         printf("%lld\n", sum[n]);

     }

 }