Two numbers are entered in from the keyboard. If the
larger number is odd, the smaller number is added to an accumulator. Then the larger number
is integer divided by two and the smaller number is doubled. Again if the larger number's
division is odd, the smaller number's doubling is added to an accumulator. lf the larger number
is even, nothing is added to the accumulator. This is repeated until the large number
sequence equals zero. The accumulator now holds the multiplication answer.
Write a program using functions, that will accomplish this task. Repeat until the user wishes to stop.
lt is not necessary to display the accumulation work as shown in the examples.
Output is to the screen and printer and should look like:
Your Name Class # Date & Time
XXXX times XXXX by conventional math = XXXXXXX
XXXX times XXXX by Zoo's method = XXXXXXX
Example 1) 75 x 23 = 1725
Larger 			Smaller 			Add to Accumulator
75			 23 				23
37 			46 				46
18 			92 				O
9 			184 				184
4 			368				 O
2 			736 				O
1 			1472 				1472
1725(TOTAL & PRODUCT)


Example 2) 
122 x 251 = 30622
Larger 		Smaller 			Add to Accumulator
251 		122 					122
125 		244 					244
62 		488 					O
31 		976 					976
15 		1952 					1952
7 		3904 					3904
3 		7808 					7808
1 		15616					 15616
30622(TOTAL & PRODUCT)

enter two numbers.
calculate the product of the two numbers.
determine which number is larger, call it A, and call the other number B.  
use an accumulator, initialzed to zero
as long as A is above zero
  if A is odd 
    add B to accumulator
  divide A by 2 using integer math (means all remainders are dropped)
  //etc

A    B    Total Accumulated so far
44 11     0
22 22     0
11 44     44
5   88     132
2   176   132
1   352   484

if(numA<numB)
   numA / 2.0
else
   accum +=numA

void getnums()
{
     
  int numA, numB, x, result;
  cout << "Your two numbers to multiply are?"<<endl;
  cin >> numA >> numB;

  if (numB < numA)
  {
    x +=numA;
    numA = numB;
    numB = x;
  }
   
  result = process( numA, numB);

  
  cout << numA << " times " << numB<< " by conventional math = " << (numA * numB) <<endl;
  cout << numA << " times " << numB<< " by Zoo's method = " << result <<endl;
  
}

if (numB < numA)
{
    // use x only inside the if-block
    int x = numA;
    numA = numB;
    numB = x;
}