This task is to find the number of 'For' loops present in the given cpp program.
--------
Question: int min(int a,int b)
{
    if(a>=b)
    return b;
    else
    return a;
}

  
  int f(int a,int b)
{ 
    int count=0;
    int k[32768]={0};
    if(a==1)
    return 1;
    else
    {
    int i,j;
    for(i=min(a,b);i>=2;i--)
    {                     
        if(a%i==0)
        k[i]=i; 
    }
    for(j=0;j<32768;j++)
    {
         if(k[j]!=0)
         {
         count=count+f(a/k[j],j);
         k[j]=0;
         }
    }
    return count; 
    }
}
main()
{
      int n,i,a,number;
      scanf("%d",&n);
      for(i=0;i<n;i++)
      {
      scanf("%d",&a);
      number=f(a,a);
      printf("%d\n",number);
      }
}


Answer: 3


Question: //n???????i?????????
int divi(int n,int i)
{
    int num=0;
    if(n%i==0)
    {
        if(n==i)      //?????????????????1 
        {
            return 1;
        }
        for(int j=2;j<=i;j++)
        {
             num=num+divi(n/i,j);   //?? 
        }
           return num;
    }
        
    else
    {
        return 0;
    }               
}

int main()
{
    int n=0;  //????? 
    int t=0;  //???? 
    int sum;  //????? 
    cin>>t;  
    for(int i=0;i<t;i++)
    {
        sum=0;
        cin>>n;
        for(int j=2;j<=n/2;j++)   //?????????1*n??? 
        {
            sum=sum+divi(n,j);
        }
        cout<<sum+1<<endl;
    }
    return 0;
}


Answer: 3


Question: int fenjie(int k, int l);
int main()
{
	int n, i;
	cin >> n;
	for(i = 1; i <= n; i++)
	{
		int a;
		cin >> a;
		cout << fenjie(a, 2) << endl;
	}
	return 0;
}

int fenjie(int k, int l)
{
	int num = 1, d, e;
	if(k < l)
	{
		return 0;
	}
	d = (int) (sqrt((double) k));
	for(e = l; e <= d; e++)
	{
		if(k % e == 0)
		{
			num += fenjie(k / e, e);
		}
	}
	return num;
}

Answer:
2