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A - Farey Sequence Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Description

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

題目大意：

就是求一個F(n)的中元素的個數。

解題思路：

首先我們分析一下題目，我們一看就是關於尤拉函式的題，就是求<=b中與b互素的個數，但是我們別忘記還得加上phi(<n)的值，所以這個問題就是求尤拉函式的和的問題，然後我們如果用正常的做法也就是暴力做的話 會超時的（我試過了/笑cry），所以我們要找到更快速的方法 ——篩法，其實這個篩法就是跟素數篩差不多的，也算是又學了一個知識點

void Init()
{
    memset(phi, 0, sizeof(phi)) ;
    for(int i=2; i<MAXN; i++)
    {
        if(!phi[i])
        {
            for(int j=i; j<MAXN;j+=i )
            {
                if(!phi[j])
                    phi[j] = j ;
                phi[j] = phi[j]/i*(i-1) ;
            }
        }
    }
}
 

Code：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
const int MAXN = 1e6+5;

int phi[MAXN];
void Init()
{
    memset(phi, 0, sizeof(phi)) ;
    for(int i=2; i<MAXN; i++)
    {
        if(!phi[i])
        {
            for(int j=i; j<MAXN;j+=i )
            {
                if(!phi[j])
                    phi[j] = j ;
                phi[j] = phi[j]/i*(i-1) ;
            }
        }
    }
}
int main()
{
    Init();
    int m;
    while(cin>>m,m)
    {
        LL sum = 0;
        for(int i=2; i<=m; i++)
            sum += phi[i];
        cout<<sum<<endl;
    }
    return 0;
}