POJ

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12802		Accepted: 4998

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

/**
          题意：求F[i] 的元素的个数，并且0<a<b<=i, gcd(a,b)=1
          做法：欧拉 用phi[n]存储小于等于n的素数的个数，然后f[i] = phi[i] + f[i]
**/
#include <iostream>
#include <string.h>
#include <cmath>
#include <algorithm>
#include <stdio.h>
#define maxn 1000100
using namespace std;
long long mmap[maxn];
int phi[maxn];
long long n;
void geteuler()
{
    memset(phi,0,sizeof(phi));
    phi[1] = 1;
    for(int i=2; i<=maxn; i++)
    {
        if(!phi[i])
        {
            for(int j=i; j<=maxn; j+=i)
            {
                if(!phi[j])
                    phi[j] = j;
                phi[j] = phi[j]/i*(i-1);
            }
        }
    }
}
int main()
{
    geteuler();
    mmap[1] = 0;
    for(int i=2; i<=maxn; i++)
    {
        mmap[i] = mmap[i-1] + phi[i];
    }
    //freopen("in.txt","r",stdin);
    while(~scanf("%lld",&n))
    {
        if(n == 0) break;
        printf("%lld
",mmap[n]);
    }
    return 0;
}