Farey Sequence （素筛欧拉函数/水）题解

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

思路：

欧拉函数模板题了吧

素筛法求得欧拉函数再加起来，记得用long long 

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<cmath>
#include<string>
#include<map>
#include<stack> 
#include<set>
#include<vector>
#include<iostream>
#include<algorithm>
#include<sstream>
#define INF 0x3f3f3f3f
#define ll long long
const int N=1e6+5;
const ll MOD=998244353;
using namespace std;
ll euler[N];
void init(){
	for(int i=0;i<N;i++) euler[i]=i;
	for(ll i=2;i<N;i++){
		if(euler[i]==i){
			for(ll j=i;j<N;j+=i){
				euler[j]=euler[j]/i*(i-1);
			}
		}
	}
	for(int i=3;i<N;i++) euler[i]+=euler[i-1];
}
int main(){
	init();
	ll n;
	while(~scanf("%lld",&n) && n){
		cout<<euler[n]<<endl;
	}
	return 0;
}