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

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

#include<stdio.h>
#include<math.h>
#define N 1000010

//欧拉函数 <模板>
int euler_phi(int n)
{
    int m = (int)sqrt(n+0.5);
    int ans = n;
    for(int i = 2; i <= m; i++)
        if(n % i == 0){//找素因子
            ans = ans/i*(i-1);
            while(n % i == 0)
                n /= i;//除尽
        }
    if(n>1)
        ans = ans/n*(n-1);
    return ans;
}

int main()
{
    int n;
    while(scanf("%d", &n) && n != 0){
        int sum = 0;
        for (int i = 2; i <= n; i ++)
            sum += euler_phi(i);
        printf("%d
", sum);
    }
    return 0;
}

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

#include<math.h>
#include<stdio.h>
#include<string.h>
#define maxn 1000010

//欧拉函数打表 <模板>
int phi[maxn];
void phi_table(int n)
{
    for(int i = 2; i <= n; i++)
        phi[i] = 0;
    phi[1] = 1;
    for(int i = 2; i <= n; i++)
        if(!phi[i])
            for(int j = i; j <= n; j += i){
                if(!phi[j])
                    phi[j] = j;
                phi[j] = phi[j]/i*(i-1);
            }
}


int main()
{
    memset(phi, 0, sizeof(phi));
    phi_table(maxn);
    int n;
    while(scanf("%d", &n) && n != 0){
        long long sum = 0;
        for(int i = 2; i <= n; i++)
            sum += phi[i];
        printf("%lld
", sum);
    }
    return 0;
}