POJ2478 Farey Sequence —— 欧拉函数

题目链接：https://vjudge.net/problem/POJ-2478

 

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17753		Accepted: 7112

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

 

 

题解：

单纯的欧拉函数。

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 1e6+10;

int euler[MAXN];
void getEuler()
{
    memset(euler, 0, sizeof(euler));
    euler[1] = 1;
    for(int i = 2; i<MAXN; i++) if(!euler[i]) {
        for(int j = i; j<MAXN; j += i)
        {
            if(!euler[j]) euler[j] = j;
            euler[j] = euler[j]/i*(i-1);
        }
    }
}

LL f[MAXN];
void init()
{
     getEuler();
    f[1] = 0;
    for(int i = 2; i<MAXN; i++)
        f[i] = f[i-1] + euler[i];
}

int main()
{
    int n;
    while(scanf("%d", &n)&&n)
        printf("%lld\n", f[n]);
}