TASK DEFINITION: This task is to find the number of 'For' loops present in the given cpp program.
PROBLEM: /*
 * h35.cpp
 * function:????
 * Created on: 2012-11-30
 * Author: sear
 */
//m????n???????????n???n??m,????n,???n??1
int f(int m, int n) {			//m???,n????????
	int i, sum = 0;				//i?????,sum???????
	if (n == 1)
		return 1;				//???????1???1
	for (sum = 0, i = m; i <= n; i++)//?????????????????m???????m???
		if (n % i == 0)
			sum += f(i, n / i);//?n???????????n??i????
	return sum;					//??sum???????
}
int main() {
	int i, n, k;				//i?????,n????????,k?????????
	for (i = 0, cin >> n; i < n; i++) {
		cin >> k;				//???????
		cout << f(2, k) << endl;//???????
	}
	return 0;				//????
}


SOLUTION: 2

PROBLEM: int fun(int m,int n)
{
    int i,sum=1;
    if(n==1)
    {
        for(i=2;i*i<=m;i++)
        {
            if(m%i==0)
            sum+=fun(m/i,i);
        }
    }
    else
    for(i=n;i*i<=m;i++)
    {
        if(m%i==0)
        {
            sum+=fun(m/i,i);
        }
    }
    return sum;
}
int main()
{
    int n,i,b;
    int a[1000];
    cin>>n;
    for(i=0;i<n;i++)
    {
        cin>>b;
        a[i]=fun(b,1);
    }
    for(i=0;i<n;i++)
    cout<<a[i]<<endl;
}


SOLUTION: 4

PROBLEM: /*
/???   10(1)
/??     ???
/??     2012?11?26?
/???   ????
*/
int main()
{
	int fenjie(int x, int y);
	int a, n, i;
	cin >> n;
	for(i = 1; i <= n; i++)
	{
		cin >> a;
cout << fenjie(a, 2) << endl;
	}
	return 0;
}
int fenjie(int x, int y)
{
	int b, i, count = 1; // ?????????
	if(x == 1)
		return 0;
	if(x == 2)
		return 1;
	b = sqrt((double) x);
	for(i = y; i <= b; i++)
	{
		if(x % i == 0)   // ?????????????????????count???
		{
			count += fenjie(x / i, i); // ????????????i??????????
		}
	}
	return count;
}

SOLUTION:
2

