【poj 2478 Farey Sequence （欧拉函数、数论）】

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9180		Accepted: 3501

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 <= 10⁶). 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

Source

POJ Contest,Author:Mathematica@ZSU

 

 

 

// Project name : 2478 ( Farey Sequence ) 
// File name    : main.cpp
// Author       : Izumu
// Date & Time  : Fri Jul 13 16:30:10 2012


#include <iostream>
#include <stdio.h>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;

#define MAXN 1000100

typedef unsigned long long int longint;

///////////////////////////////////////////////////////////////////
bool prime[MAXN];
///////////////////////////////////////////////////////////////////
void CInitPrime()
{
    for (int i = 0; i < MAXN; i++)
    {
        prime[i] = true;
    }

    int current_prime = 2;
    for (; current_prime * current_prime <= MAXN; current_prime++)
    {
        if (prime[current_prime] == true)
        {
            for (int i = current_prime * current_prime; i < MAXN; i = i + current_prime)
            {
                prime[i] = false;
            }
        }
    }
}
///////////////////////////////////////////////////////////////////
longint phi[MAXN];
///////////////////////////////////////////////////////////////////
void CInitPhi()
{
    for (int i = 1; i < MAXN; i++)
    {
        phi[i] = i;
    }

    for (int i = 2; i < MAXN; i++)
    {
        if (prime[i])
        {
            for (int j = i; j < MAXN; j += i)
            {
                phi[j] = phi[j] / i * (i - 1);
            }
        }
    }
}
///////////////////////////////////////////////////////////////////
int main()
{
    CInitPrime();
    CInitPhi();
    int index;
    while (cin >> index && index)
    {
        longint sum = 0;
        for (int i = 2; i <= index; i++)
        {
            sum += phi[i];
        }
        cout << sum << endl;
    }
    return 0;
}

// end 
// ism