1) 15 points 

   a) (float) num

   b) long double percentage;

   c) short int foo;
      unsigned short int foo;	and
      char foo;			are also acceptable

   d) 10

   e) 4

2) 34 points

a)  12 5 2
b)  5 9 13 17 21 
c)  0 2 1 3 2 5 5
d)  2 3 2 7 2 11 
e)  25 16 9 Yes


3) 15 points
You can use it to determine whether three numbers are the lengths of
the sides of a right triagle.


4) 36 points

a) 

int order(char arr[], int len);

int numordered(char arr[][80], int lens[], int rows);


b) 
   numordered(words, lengths, 30);

c) 

int order1 (char a[], int len) {

   int i, result = 1;

   for (i = 0; i < len - 1; i++)
      if (a[i] > a[i+1])
         result = 0;

   return result;

}

int order2 (char a[], int len) {

/* while loop version (more efficient) */

   int i = 0, ordered = 1;

   while (ordered && i < len - 1)
   {
      if (a[i] > a[i+1])
         ordered = 0;
      i++;
   }

   return ordered;

}

int order3 (char a[], int len) {

   int i, result = 0, counter = 0;

/* counter version */

   for (i = 0; i < len - 1; i++)
      if (a[i] <= a[i+1])
         counter++;
   if (counter == len - 1)
      result = 1;

   return result;

}



int numordered(char arr[][80], int lens[], int rows)
{
   int i, count = 0;

   for(i = 0; i < rows; i ++)
      if(order(arr[i], lens[i]))
	 count ++;
   return count;
}

OR

int numordered2(char arr[][80], int lens[], int rows)
{
   int i, count = 0, j;
   char tmparray[80];

   for(i = 0; i < rows; i ++)
   {
      for(j = 0; j < 80; j++)
         tmparray[j] = arr[i][j];
      if(order(tmparray, lens[i]))
	 count ++;
   return count;
}