This task is to find the number of 'For' loops present in the given cpp program.

//***************************************
//* ?????2010?12?01?            *
//* ?????????                  *
//* ?????         ???1000012902 *
//***************************************
int factors(int, int, int);                    //????factors?????????????
int main() 
{
	int num, n, i, j, b[10000];
	cin >> n;                                  //??????
	for (i = 0; i < n; i++) 
	{
		cin >> num;
		int part = 0;                          //??????????
		int t = num;                           //????

		//??????1????????????
		for (j = 2; j <= num; j++)
		{
			while (t % j == 0)
			{
				t /= j;
				part++;
			}
		}
		b[i] = factors(1, part, num);          //?????????????, ??????????
	}

	//?????????
	for (i = 0; i < n; i++)
	{
		cout << b[i] << endl;
	}
	return 0;
}

//factor1????????????????????factor1, part??????????????????????????, num???????? 
int factors(int factor1, int part, int num)
{
	int sum = 0;                               //????????0, ???0
	if (num < factor1)                         //?????????????????, ???0?????????????
		return 0;
	if (part == 1)                             //?????????, ??1??????????
		return 1;

	//???????????????????????????????????????????, ????????
	for (int i = factor1; i <= num; i++)
		if (num % i == 0)
			sum = sum + factors(i, part - 1, num / i);
	return sum;
}
4

int Digui(int a,int min)
{
	int i,result=1;
	if(a<min)
	{
		return 0;
	}
	for(i=min;i<a;i++)
	{
		if(a%i==0)
		{
			result+=Digui(a/i,i);
		}
	}
	return(result);
}
int main()
{
	int a[1000],n,i;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
	}
	for(i=0;i<n;i++)
	{
		printf("%d\n",Digui(a[i],2));
	}
	return 0;
}



3


/**
* @author ??
* @date 2010-12-1
* @description
* ??????:???? 
*/
int sum;
void calculate(int, int);


int main()
{
	int n;
	cin >> n;
	for (int i = 0; i < n;i++)
	{
		int m;
		sum  = 1;
		cin >> m;
		calculate(m, 2);
		cout << sum << endl;
	}
	return 0;
}
void calculate(int m,int begin)
{
	int i;
	int n = sqrt((double)m);
	for ( i = begin; i <= n; i++ )
	{
		if ( m % i == 0)
		{
			sum++;
			calculate(m / i,i);
		}
	}
}

2