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+;
 const int MAXN = 1e6+;

 int euler[MAXN];
 void getEuler()
 {
     memset(euler, , sizeof(euler));
     euler[] = ;
     for(int i = ; 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-);
         }
     }
 }

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

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