Woj problem 1420-Visible point

Problem 1420-Visible point Time Limit: 3000MS Memory Limit: 65536KB Difficulty: 3
Total Submit: 539 Accepted: 113 Special Judge: No Description plane has m*n on the whole point, their coordinates (x, y) meet 1<=x<=m, 1<=y<=n, x, y are integers. The number of points that can be seen from the origin (if there is no other point on the line to the origin, the point can be seen by the origin).
Input first line a number T (1&LT;=T&LT;=15), which represents the number of groups of data
One row per group of data, two digits per line m,n (0<=m, n<=50000)
Output for each set of data, the total number of points that can be seen by the origin.
Sample Input 2
1 1
2 3
Sample Output 1

5

#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
using namespace Std;

int n,m;
int Ans;

Vector<int> F;

void Dfs (int step,int frac,int Tot)
{
if (Step >= f.size ())
{
if (tot&1)
Ans-= (Long Long) (M/FRAC);
Else
Ans + = (long Long) (M/FRAC);
Return
}
Dfs (step+1,frac*f[step],tot+1);
Dfs (Step+1,frac,tot);
}
void Count (int Num)
{
int Cur = Num;
F.clear ();
for (int i=2;i*i<=num;i++)
if (cur%i = = 0)
{
F.push_back (i);
while (cur%i = = 0)
Cur/= i;
}
if (Cur > 1) f.push_back (Cur);
Dfs (0,1,0);
}

int main ()
{
int n;
CIN >> N;
while (n>0)
{
Ans = 0;
CIN >>M>>N;
for (int i=1;i<=n;i++)
Count (i);
cout<<ans<< ' \ n ';
N--;
}
}