TASK DEFINITION: This task is to find the number of 'For' loops present in the given cpp program.
PROBLEM: int n,i;//????  n ???????
int f(int a,int min){      //??????f  ?????a n
if(a < min){     //?????a ??n
return 0;     //????0?   ???????????
}
int result = 1;       //???????  ???1
for(int i = min;i<a;i++){   //for?? ? ??????a
if(a % i == 0){   //?a???i???
result += f(a/i,i);  //???? ??result
}
}
return result; //??????
}
      int main(){  //?????
    	  cin>>n;//??????
    	  int a[n];
    	  for(i=0;i<n;i++){   //????
    		  cin>>a[i];
    	  }
    	  for(i=0;i<n;i++){
    	  cout<<f(a[i],2)<<endl;  //??????????  ????
    	  }
    	  return 0;
      }

SOLUTION: 3

PROBLEM: int factoring(int a,int b)
{
    int i,kind=0;
	if(a==1)
	{
	    kind=1;
	}

	else
	{
	    for(i=b;i<=a;i++)
		{
		    if(a%i==0)
			{
	            kind=factoring(a/i,i)+kind;
			}
		}
	}
	return kind;
}

int main()
{
	int n,i,j,a[30000]={0},b[30000]={0};
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
		b[i]=factoring(a[i],2);
	}
	for(i=0;i<n;i++)
	{
		printf("%d\n",b[i]);
	}
	return 0;
}

SOLUTION: 3

PROBLEM: int fj(int a, int b);

int main()
{
	int n, i, a;
	cin >> n;
	for(i = 0; i < n; i++)
	{
		cin >> a;
		cout << fj(a, 2) << endl;
	}

	return 0;
}
int fj(int a, int b)
{
	int j, x, sum = 1, f = 0;
	x = sqrt((double) a);
	if((a == 1) || (a == 2) || (a == 3))
		return 1;
	else
	{
	    for(j = b; j <= x; j++)
	    {
		    if(a % j == 0)
			{
			    sum += fj(a / j, j);
				f = 1;
			}
    	}
		if (f == 0)
	        return 1;
		else
			return sum;
	}
}

SOLUTION:
2