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 <= 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

打表使用euler函数公式，注意当中巧妙的使用筛子的方法。

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;
}

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