POJ 2478 Farey Sequence

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

Sample Output

1
3
5
9


欧拉函数的应用

很简单的一道题，但是直接使用欧拉函数会TLE

 1 #include<stdio.h>
 2 #include<math.h>
 3 #define N 1000010
 4 
 5 //欧拉函数 <模板>
 6 int euler_phi(int n)
 7 {
 8     int m = (int)sqrt(n+0.5);
 9     int ans = n;
10     for(int i = 2; i <= m; i++)
11         if(n % i == 0){//找素因子
12             ans = ans/i*(i-1);
13             while(n % i == 0)
14                 n /= i;//除尽
15         }
16     if(n>1)
17         ans = ans/n*(n-1);
18     return ans;
19 }
20 
21 int main()
22 {
23     int n;
24     while(scanf("%d", &n) && n != 0){
25         int sum = 0;
26         for (int i = 2; i <= n; i ++)
27             sum += euler_phi(i);
28         printf("%d
", sum);
29     }
30     return 0;
31 }


于是，我们需要进行一些改进，欧拉函数可以写成递归形式，类似于记忆化搜索，我们不需要每次都重复计算,也就是所谓的欧拉函数打表
其中需要注意的一点就是，当 n 的值比较大的时候， sum 会超出 int 的范围（当输入 n 的值为100000时就会发现输出为负值），所以要改成long long 类型或者 _int64 类型

 1 #include<math.h>
 2 #include<stdio.h>
 3 #include<string.h>
 4 #define maxn 1000010
 5 
 6 //欧拉函数打表 <模板>
 7 int phi[maxn];
 8 void phi_table(int n)
 9 {
10     for(int i = 2; i <= n; i++)
11         phi[i] = 0;
12     phi[1] = 1;
13     for(int i = 2; i <= n; i++)
14         if(!phi[i])
15             for(int j = i; j <= n; j += i){
16                 if(!phi[j])
17                     phi[j] = j;
18                 phi[j] = phi[j]/i*(i-1);
19             }
20 }
21 
22 
23 int main()
24 {
25     memset(phi, 0, sizeof(phi));
26     phi_table(maxn);
27     int n;
28     while(scanf("%d", &n) && n != 0){
29         long long sum = 0;
30         for(int i = 2; i <= n; i++)
31             sum += phi[i];
32         printf("%lld
", sum);
33     }
34     return 0;
35 }