CSU 1131: Nim-B* Sum (求B進位制的異或和)
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).