POJ2478

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

这道题的题意是找所有比n小的互质的a和b，所以我套了欧拉公式的模板。主要是遍历所有比n小的数，设为x，再把所有phi[x]加起来就是答案。
欧拉公式：phi[n]，比n小，且与n互质的正整数个数。

for (int i = 2; i <= 1000000; i++) {
        if (!phi [i] ) {
            for (int j = i ; j <= 1000000 ; j += i) {
                if (!phi [j] )
                    phi[j] = j ;
                phi[j] = phi [j] / i * (i - 1) ;
            }
        }
    }

上面的是欧拉公式的一种实现方式。

#include <iostream>
#include <cstdio>

using namespace std;
const int maxn = 1000000 + 5;
int phi[maxn] = {0};

int main()
{
    #ifndef ONLINE_JUDGE
    freopen ("in.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    int n;
    for (int i = 2; i <= 1000000; i++) {
        if (!phi [i] ) {
            for (int j = i ; j <= 1000000 ; j += i) {
                if (!phi [j] )
                    phi[j] = j ;
                phi[j] = phi [j] / i * (i - 1) ;
            }
        }
    }
    while (cin >> n) {
        if (n == 0) break;
        long long N = 0;//long long存储
        for (int i = 2; i <= n; i++) {
            N += phi[i];
        }
        cout << N << endl;
        //cout << endl;
    }
    return 0;
}