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.
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
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<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;
}
