A - Farey Sequence——（筛法求欧拉函数）-白红宇

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

Sample Output

题目大意：

就是求一个F(n)的中元素的个数。

解题思路：

首先我们分析一下题目，我们一看就是关于欧拉函数的题，就是求<=b中与b互素的个数，但是我们别忘记还得加上phi(<n)的值，所以这个问题就是求欧拉函数的和的问题，然后我们如果用正常的做法也就是暴力做的话会超时的（我试过了/笑cry），所以我们要找到更快速的方法 ——筛法，其实这个筛法就是跟素数筛差不多的，也算是又学了一个知识点

void Init(){    memset(phi, 0, sizeof(phi)) ;    for(int i=2; i

Code：

#include 
      
       #include 
       
        #include 
        
         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
         
          >m,m)    {        LL sum = 0;        for(int i=2; i<=m; i++)            sum += phi[i];        cout<
          
           <