原【欧拉函数模板题】【前缀和】FareySequence

原【欧拉函数模板题】【前缀和】FareySequence

Problem 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<sup>6</sup>). 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. <b

Sample Input

2

3

4

5

0

 

Sample Output

1

3

5

9

思路:由a / b,gcd(a,b)=1知,当前f(n)是在f(n - 1)的基础上加上以n为分母,与n互质的数为分子的分数,所以f(n)比f(n - 1)增加了1~n内与n互质的数的个数,现在题意就很明显了,就是要求1-N的欧拉函数之和。

 

#include <iostream>
using namespace std;
typedef long long ll;
const int MAX=1000005;
int oula[MAX];
void init()
{
	oula[1]=1;
	for(int i=2;i<MAX;i++)
		oula[i]=i;
	for(int i=2;i<MAX;i++)
		if(oula[i]==i)
			for(int j=i;j<MAX;j+=i)
				oula[j]=oula[j]/i*(i-1);
}
int main()
{
	ll n,ans=0;
	init();
	while(cin>>n&&n)
	{
		ans=0;
		for(int i=2;i<=n;i++)
			ans+=oula[i];
		cout<<ans<<endl;
	}
	return 0;
}

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