Description

The game of NIM is played with any number of piles of objects with any number of objects in each 
pile.  At each turn, a player takes one or more (up to all) objects from one pile.  In the normal form of 
the game, the player who takes the last object is the winner.  There is a well-known strategy for this 
game based on the nim-2 sum. 
 
The Nim-B sum (nim sum base B) of two non-negative integers X and Y (written NimSum(B, X, Y)) 
is computed as follows: 
 
1)  Write each of X and Y in base B. 
2)  Each digit in base B of the Nim-B sum is the sum modulo B of the corresponding digits in the 
base B representation of X and Y. 
 
For example: 
NimSum(2, 123, 456) = 1111011 ¤   111001000 = 110110011 = 435 
NimSum(3, 123, 456) = 11120 ¤   121220 = 102010 = 300 
NimSum(4, 123, 456) = 1323 ¤   13020 = 10303 = 307 
 
The strategy for normal form Nim is to compute the Nim-2 sum T of the sizes of all piles.  If at any 
time, you end your turn with T = 0, you are guaranteed a WIN.  Any opponent move must leave T not 
0 and there is always a move to get T back to 0.  This is done by computing 
NimSum(2, T, PS)for each pile; if this is less than the pile size (PS), compute the difference 
between the PS and the Nim-2 sum and remove it from that pile as your next move. 
 
Write a program to compute NimSum(B, X, Y).