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 <= 10 ⁶). 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


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\n", 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\n", sum);

33     }

34     return 0;

35 }