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


Question: int main(){
	int number(int n, int low);
	int n, i, data, ans;
	scanf("%d", &n);
	for (i=1; i<=n; i++)
	{
		scanf("%d", &data);
		ans = number(data, 2);
		printf("%d\n", ans);
	}
}
int number(int n, int low)
{
	int j, ans;
	ans = 0;
	for (j = low ; j * j<= n; j++)
	{
		if (n % j == 0) 
			ans = ans + number(n / j, j);
	}
	ans = ans + 1;
	return ans;
}

Answer: 2


Question: int n,m;
int sum(int x,int y)
{
   int c,i;
   if(x>sqrt(y))
       c=0; 
   else
   {
       if(y%x==0)
       {
           for(i=x,c=1;i<=(y/2)+1;i++)
           {
               if((y/x)%i==0)
                   c=c+sum(i,(y/x));
           }
       }
       else
           c=0;
   } 
   return(c);
}
main()
{
    int i,k,j;
    scanf("%d",&n);
    for(j=1;j<=n;j++)
    {
        k=0;
        scanf("%d",&m);
        for(i=2;i<=(m/2)+1;i++)
            k=k+sum(i,m);
            printf("%d\n",k+1);   
    }
    }
    

Answer:
3


