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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}
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.
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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