TASK DEFINITION: This task is to find the number of 'For' loops present in the given cpp program.
PROBLEM: int f(int x,int y)
{
	int a,n=0;
	for(a=y;a<=sqrt(x);a++)
	{
		if(x%a==0)
		{
			n++;
			n+=f(x/a,a);
		}	
	}
	return n;
}
int main()
{
	int m,i;
	scanf("%d",&m);
	int x[m],n[m];
	for(i=0;i<m;i++)
	{	
		scanf("%d",&x[i]);
		n[i]=1+f(x[i],2);
	}
	for(i=0;i<m;i++)
	{
		printf("%d\n",n[i]);	
	}
	return 0;
}


SOLUTION: 3

PROBLEM: /*
 *  1000012850_34.1.cpp
 *  ????????????a??????????????? ?????????
 *  Created on: 2010-11-27
 *      Author: william
 */
int resolution_into_factors(int j,int i)
{
	int m, sum=0, k=0;
	for(m=j ;m<=sqrt(i); m++)
	{
		if(i%m==0)
		{
            sum=resolution_into_factors(m, i/m)+sum+1;
			k++;
		}
	}
	if(k==0) return 0;
    return sum;
}
int main()
{
    int n, k, i;
    cin>>n;
    for(k=1; k<=n; k++)
    {
    	cin>>i;
    	cout<<resolution_into_factors(2,i)+1<<endl;
    }
    return 0;
}

SOLUTION: 2

PROBLEM: void account (int ,int);
int p;
int main()
{
	int m;
	int n;
	cin>>n;
	while(n>0)
	{
		p=0;
    cin>>m;
account(m,1);
cout<<p<<endl;
n--;
	}
return 0;
}
void account (int m,int i)
{
if(m==1)
    p++;
else for(int j=2;j<=m;j++)
	{
	if((m%j==0)&&(j>=i))
     account(m/j,j);
	}
}

SOLUTION:
1