Getting started with the game (ACM game combination)

Miyu original, post Please note: Reprinted from__________ White House

Find a balance (also known as a defeat, Singular situation), (YES: any non-equilibrium state can be changed to an equilibrium state after an operation)

(I)Bashi game(Bash game ):

only a bunch of n items , two people take things from these items in turn , specify at least one , up to m . winners .

N = (m + 1) R + S, (R is any natural number, S ≤ m), that is, N % (m + 1 )! = 0, the first accessor must win

(II)Weizoboyi(Wythoff game ):

There are two heaps of items,Two people take the same number of items from one or both of them in turn.,Requires at least one,Unlimited,The final winner.

(AK, BK) (Ak ≤ BK, K = 0, 1, 2,..., n) indicatesSingular situation

Method:

Ak = [K (1 + √ 5)/2], BK = ak + K (k = 0, 1, 2,..., n square brackets indicate the entire function)

Judgment:

Gold = (1 + SQRT (5.0)/2.0;

1) Assume that (a, B) is a singular situation of the K type (k =, 2...) So k = B-;

2) judge its A = (INT) (K * gold). If it is equal, it is a strange situation.

(Note: Use appropriate methods,The non-singular situation can be changed to a singular situation..

Assume that the situation is (a, B)

If B = A, a object is taken from both heaps and changed to a singular situation (0, 0 );

1.If a = ak,

1.1B> BK, then, the objects B-BK are taken away, that is, the singular situation (AK, BK );

1.2B <BK takes Ak-A [B-Ak] objects from the two heaps at the same time and changes to a singular situation (A [B-Ak], A [B-Ak] + B-ak );

2If a = BK,

2.1B> AK, the excess number of B-ak is removed from the second heap.

2.2B <AK, if B = AJ (j <k) takes the excess number A-BJ from the first heap; (a> BJ)

If B = BJ (j <K) is removed from the first heap a-AJ; (a> AJ)

)

Example:PKU 1067

( 3 ) nimboi (nimm game):

Yes n items in the heap , two people take multiple items from a pile in turn , specify at least one , unlimited , final winner .

AnySingular situation(A1, A2 ,... , An) all have A1 (+) A2 (+ )... (+) An = 0. (+) is by bit ^)

About Nim games:

(1) There are two players;

(2) There are three groups of playing cards (for example, 5, 7, and 9 cards );

    The two sides of the game operate in turn;

(3) Each player's operation is to select a pile of cards and take any cards from them;

    The winner of the last shot;

Definition: Suppose (XM · X0) 2 and (ym · y0) 2 Nim-sum is (ZM · z0) 2, then we represent (XM · X0) 2 values (ym · y0) 2 = (ZM · z0) 2,
Here, zk = XK + YK (mod 2) (k = 0... M ).

Theorem 1 of Nim games:

For a nim game location (x1, x2, X3), when and only when the Nim-sum of each of its parts is equal to 0 (that is, X1 ready X2 ready X3 = 0) (x1, x2, X3 do an exception or operation ^), it is currently in a bid.

 Theorem 1 applies to more heap cases!

Example:PKU 2234

Example:HDU 1730

Example:PKU 1740

Example:PKU 1704

Example:PKU 1082 (Massive Analysis...The conclusion is simple. It can also be simulated based on simple inferences.)