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

long long ans=0;
for(i=2;i<=n;i++)
ans+=oula[i];


#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int oula[1000005];

int main()
{
int n;
int i,j;
memset(oula,0,sizeof(oula));
for(i=2;i<=1000000;i++)
{
if(!oula[i])
{
for(j=i;j<=1000000;j+=i)
{
if(!oula[j])
oula[j]=j;
oula[j]=oula[j]/i*(i-1);
}
}
}
while(cin>>n,n)
{
long long ans=0;
for(i=2;i<=n;i++)
ans+=oula[i];
cout<<ans<<endl;
}
return 0;
}

0
Posted in ACM