POJ2478(SummerTrainingDay04-E 欧拉函数)

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16927		Accepted: 6764

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

 

求1-n的欧拉函数和即可。

 //2017-08-04
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>

 using namespace std;

 const int N = ;
 int phi[N],prime[N],tot;
 long long ans[N];
 bool book[N];

 void getphi()//线性欧拉函数筛
 {
    int i,j;
    phi[]=;
    for(i=;i<=N;i++)//相当于分解质因式的逆过程
    {
        if(!book[i])
        {
             prime[++tot]=i;//筛素数的时候首先会判断i是否是素数。
             phi[i]=i-;//当 i 是素数时 phi[i]=i-1
         }
        for(j=;j<=tot;j++)
        {
           if(i*prime[j]>N)  break;
           book[i*prime[j]]=;//确定i*prime[j]不是素数
           if(i%prime[j]==)//接着我们会看prime[j]是否是i的约数
           {
              phi[i*prime[j]]=phi[i]*prime[j];break;
           }
           else  phi[i*prime[j]]=phi[i]*(prime[j]-);//其实这里prime[j]-1就是phi[prime[j]]，利用了欧拉函数的积性
        }
    }
 }

 int main()
 {
     int n;
     getphi();
     ans[] = ;
     for(int i = ; i < N; i++){
         ans[i] = ans[i-]+phi[i];
     }
     while(scanf("%d", &n) && n){
         printf("%lld\n", ans[n]);
     }

     return ;
 }