This task is to find the number of 'For' loops present in the given cpp program.
--------
Question: /*????
 * divide.cpp
 *
 *  Created on: 2011-12-1
 *      Author: WJ
 */

int sum=1;                            //?????1?????
void f(int b,int a)                        //??????????????
{
	for(int i=b;i<=a;i++){           //????????
	if((a%i==0)&&(i<=a/i)){
		sum=sum+1;                   //???1
		f(i,a/i);                         //???????????
	}
	if(i>a/i)  break;
	}
}
int main()
{
	int a,n;
	cin>>n;
	while(n>0){                //??????
		cin>>a;
		sum=1;
		f(2,a);                  //????
		cout<<sum<<endl;         //????
		n--;
	}
	return 0;
}


Answer: 1


Question: int number=0;
int main()
{
	void f(int first,int m);
	int n,i,num;
	cin>>n;
	for(i=0;i<n;i++)
	{
		cin>>num;
		f(2,num);
		cout<<number/2<<endl;
		number=0;
	}
	return 0;
}
void f(int first,int m)
{
	int i;
	for(i=first;i<=m;i++)
	{
		
		if(m%i==0)
		{
			m=m/i; 
			f(i,m);	
			if(m!=1)
			m=m*i;
		
		}
	}
	if(m==1)
		number+=1;  
}


Answer: 2


Question: 
/*
????
??
	?????????????
	f[i,j]????i????j?????????
	??????????
	i?j????N?????O(sqrt(n))???
?????
	?????O(n^(3/2))
	?????O(n)
*/
int main()
{
	int f[200][200] = {0};
	int div[40000] = {0};
	int divth[200] = {0};
	int kase = 0;
	cin >> kase;
	for (; kase > 0; kase --)
	{
		int n = 0;
		cin >> n;
		memset(f,0,sizeof(f));
		memset(div,0,sizeof(div));
		memset(divth,0,sizeof(divth));
		int m = 0;
		for (int i = 1; i <= n; i ++) 
			if (n % i == 0) 
			{
				div[i] = ++m;
				divth[m] = i;
			}
		f[1][1] = 1;
		for (int i = 2; i <= m; i++)
			for (int j = 2; j <= i; j++)
				for (int k = j; k > 1; k--)
					if (divth[i] % divth[k] == 0)
					{
						int ii = divth[i] / divth[k];
						int jj = k;
						if (div[ii] < jj) jj = div[ii];
						f[i][j] += f[div[ii]][jj];
					}
		cout << f[m][m] << endl;
	}
	return 0;
}	

Answer:
5