POJ2478(SummerTrainingDay04-E 欧拉函数)

Posted 2020-10-01 Penn000

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:[email protected]

 

求1-n的欧拉函数和即可。

 1 //2017-08-04
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 
 7 using namespace std;
 8 
 9 const int N = 1000010;
10 int phi[N],prime[N],tot;
11 long long ans[N];
12 bool book[N];
13 
14 void getphi()//线性欧拉函数筛
15 {    
16    int i,j;    
17    phi[1]=1;    
18    for(i=2;i<=N;i++)//相当于分解质因式的逆过程    
19    {    
20        if(!book[i])
21        {    
22             prime[++tot]=i;//筛素数的时候首先会判断i是否是素数。    
23             phi[i]=i-1;//当 i 是素数时 phi[i]=i-1    
24         }    
25        for(j=1;j<=tot;j++)    
26        {    
27           if(i*prime[j]>N)  break;    
28           book[i*prime[j]]=1;//确定i*prime[j]不是素数    
29           if(i%prime[j]==0)//接着我们会看prime[j]是否是i的约数    
30           {    
31              phi[i*prime[j]]=phi[i]*prime[j];break;    
32           }    
33           else  phi[i*prime[j]]=phi[i]*(prime[j]-1);//其实这里prime[j]-1就是phi[prime[j]]，利用了欧拉函数的积性    
34        }    
35    }    
36 }
37 
38 int main()
39 {        
40     int n; 
41     getphi();
42     ans[2] = 1;
43     for(int i = 3; i < N; i++){
44         ans[i] = ans[i-1]+phi[i];
45     }
46     while(scanf("%d", &n) && n){
47         printf("%lld\n", ans[n]);
48     }
49 
50     return 0;
51 }