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

Sample Output

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;

}

巴特西

Farey Sequence （素筛欧拉函数/水）题解

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