Given the task definition, example input & output, solve the new input case.
This task is to find the number of 'For' loops present in the given cpp program.
Example: main()
{
      float a[4];
      float k;
      float c;
      int i;
      for(i=0;i<4;i++)
      {
                      scanf("%f",&a[i]);
      }
      scanf("%f",&c);
      c=2*3.1415926*c/360;
      k=(a[0]+a[1]+a[2]+a[3])/2;
      double s,sq;
      sq=(k-a[0])*(k-a[1])*(k-a[2])*(k-a[3]) - a[0]*a[1]*a[2]*a[3]*(double)cos(c/2)*(double)cos(c/2);
    
      if(sq<0)
      printf("Invalid input");
      else
      {
          s=sqrt(sq);
          printf("%.4f",s);
      }
    
}

Output: 1
The number of 'for' loops in the code is given by the number of 'for' string present in the code. Since we can find the exact number of for loops, this is a good example.

New input case for you: /*
 *????HW7_10.cpp
 *??????1100019007?
 *???2012-11-14 
 *???1090 ???? 
 */ 
int factorize(int m,int k)
{
    int c=1;//a=a??? 
    if(m<4)return 1;//?m??4??1 
    for(int i=k;i<=sqrt(m);i++)if(!(m%i))c+=factorize(m/i,k>i?k:i);//????????????????an>an-1,???k???
    return c;//????? 
}
int main(int argc, char *argv[]){
    int n,m;
    cin>>n;
    while(n--){
        cin>>m;//????
        cout<<factorize(m,2)<<endl;//????
    }
    return 0;}

Output:
1