given two positve integers between 0 and 9 inclusively: a1 and a 2 , such that a1 >=  a2.

1. IF a2 >= 10 - a1 THEN a1 + a2 >= 10
    i = 10 - a1
    j = a2 - i
    a1 + a2 = 10 + j

2. 9 + n = 10 + (n - 1) where 0 < n <= 9

3. 0 + n = n where n is any real number

examples:

1.  5 + 7 = ?
    a1 = 7 a2 = 5
    5 >= 10 - 7 --> 5 >= 3 True
    i = 10 - 7 --> i = 3
    j = 5 - 3 --> j = 2
    5 + 7 = 10 + 2 --> 12

2. 6 + 9 = ?
   10 + (6 - 1) --> 10 + 5 --> 15

3. 0 + 4 = 4 duh!

/******************************************************************************\
|programmer: Christopher
|conact: mrynit@yahoo.com
|date made: 5-31-06
|last mod: 5-31-06
|compiler: Borland C++ Compiler 5.5
|OS: XPpro SP2
|CPU: AMD Athlon 64 +3000
|
|goal: the addtion of one digit positive integers by specific functions.
|
|license: free for personal and eduactional uses. please give acknowledgments 
|to me.
| 
\******************************************************************************/

#include <iostream>
#include <ctime>	 // For time()
#include <cstdlib>  // For srand() and rand()
using namespace std;

int main()
{
	//varriable declaration
	int n1,n2,a1,a2,sum,i,count = 0;
	
	//generate two random integer between zero and nice inclusively
	srand( (unsigned)time( NULL) ); //get random seed from cpu clock
	
	n1 = rand()%10;
	n2 = rand()%10;

	//sort n1 and n2 so that a1 is the largest of n1 and n2. a2 is the least.
	if (n1 >= n2)
	{
		a1 = n1;
		a2 = n2;
	}
	else
	{
		a1 = n2;
		a2 = n1;
	}
	
	//if both a1 and a2 are zero then sum is zero
	if (a1 == 0)
	{
		if (a2 == 0)
			sum = 0;
	}
	
	//if a2 is zero then sum is a1
	if (a2 == 0)
		sum = a1;
	
	//if a1 is nine and a2 is not zero then sum is a2 minus one
	if (a1 == 9)
	{
		if (a2 != 0)
		{
			sum = (a2 - 1);
			count++;//carry the 10
		}
	}
	
	//if a2 is greater than or equal to ten minus a1, which equals i, 
	//then a2 minus i is the sum.
	//if a2 is not greater than or equal to ten minus a1
	//then sum is a1 plus a2.
	
	//the above logic takes care of all nine plus any number combinations
	//thus if a1 is nine this block of code does not need to be run.
	//this also prevents the carry of 10 being counted twice.
	if (a1 != 9)
	{
		i = (10 - a1);
		if (a2 >= i)
		{
			sum = a2 - i;
			count++;//carry the 10
		}
		else
			sum = a1 + a2;
	}
	

	//display the sum of a1 and a2.
	cout << a1 << " + " << a2 << " = " << (sum + (count * 10));
	//(sum + (count * 10)) will be changed when programm handdles larger numbers.	
		
  return 0;
}