Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14230		Accepted: 5624

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
题意：问满足a/b with 0 < a < b <= n and gcd(a,b) = 1,的数对有多少个;
思路：dp[i]=dp[i-1]+n的欧拉函数;
AC代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int N=1e6+;

long long dp[N],a[N];

int get_a()

{

    memset(a,,sizeof(a));

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

    {

        if(!a[i])

        {

            for(int j=i;j<N;j+=i)

            {

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

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

            }

        }

    }

}

int fun()

{

    get_a();

    dp[]=;

    dp[]=;

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

    {

        dp[i]=dp[i-]+a[i];

    }

}

int main()

{

    int n;

    fun();

    while()

    {

        scanf("%d",&n);

        if(!n)break;

        cout<<dp[n]<<"\n";

    }

    return ;

}