Y - Farey Sequence
虽然菜，但不甘于菜;
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
一个欧拉函数模板题，求phi数组，之前还一直好奇为什么不用把素数都筛选出来，然后看到有2个if判断的时候知道啦，
公式就是的连乘n(1-1/pi)；
这个模板在刘汝佳的算法竞赛入门数论基础的计数那个板块；
刚刚才知道可以插入代码，23333；
难得的一发ac；贴下代码；
#include<cstdio>
#include<cmath>
#include<iostream>
#include<cstring>
using namespace std;
const int N=1e6+5;
typedef long long ll;
ll phi[N];
void Find_phi()
{
memset(phi,0,sizeof(phi));
for(ll i=2;i<=N;i++)
if(!phi[i])
for(ll j=i;j<N;j+=i)
{
if(!phi[j]) phi[j]=j;
phi[j]=phi[j]/i*(i-1);
}
}
int main()
{
//freopen("input.txt","r",stdin);
int n;
Find_phi();
while(~scanf("%d",&n)&&n)
{
ll sum=0;
for(ll i=1;i<=n;i++)
sum+=phi[i];
printf("%lld\n",sum);
}
return 0;
}
