POJ 2478 Farey Sequence

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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

23450

Sample Output

1359

打表使用euler函數公式，注意其中巧妙的使用篩子的方法。

const int MAX_SZIE = 1000001;__int64 phi[MAX_SZIE];void eulerPhi(){memset (phi, 0, sizeof(phi));for (int i = 2; i < MAX_SZIE; i++){if (!phi[i]){for (int j = i; j < MAX_SZIE; j += i){if (!phi[j]) phi[j] = j;phi[j] = phi[j] / i * (i - 1);}}}for (int i = 3; i < MAX_SZIE; i++){phi[i] += phi[i-1];}}int main(){eulerPhi();int n;while (scanf("%d", &n) && n){printf("%lld\n", phi[n]);}return 0;}