Game-nim Games

1069 Nim Games

Base time limit: 1 seconds space limit: 131072 KB score: 0 Difficulty: Basic problem

There are n heaps of stones. A B Two people take turns, a take first. Each time can only from a bunch of a few, can be a bunch of all take away, but not to take, the last 1 stones to win the people. Suppose a B is very clever, there is no mistake in the process of taking the stone. Give the number of N and each heap of stones, and ask who can win the game at the end. For example: 3 piles of gravel, 1 capsules per heap. A take 1, B take 1, at this time there are 1 piles left, so a can get the last 1 stones.

Input

Line 1th: A number n, indicating that there are n heap of stones. (1 <= n <= 1000) 2-n + 1 rows: n heap of stones. (1 <= a[i] <= 10^9)

Output

If a WINS output A, if B wins output B.

Input example

3 1 1 1

Output example

A

"Idea handling from http://www.cnblogs.com/easonliu/p/4472541.html"

This time we are talking about an old and classic game problem: the Nim game.

The Nim game is the classic Fair Combo Game (ICG), for the ICG game we have the following definitions:
1, two players;
2. Two players take turns, each action can choose one in the limited legal operation set;
3, any kind of game possible situation (position), the legal operation set only depends on the situation itself; the change of situation is called "move".
4. If the player's legal operation set is empty when the player is in the turn, the contestant will be awarded a negative.

For the third, we have a further definition of position, we will position into two categories:
P-position: In the current situation, the initiator will be defeated.
N-position: In the current situation, the initiator win.

They have the following properties:
1. The situation that the legal operation set is empty is p-position;
2. The situation that can be moved to P-position is n-position;
3. All movements can only reach the n-position situation is p-position.

In this game, we already know that a[] = {0,0,..., 0} is in P situation, then we can deduce all possible situations by reverse enumeration, the total number of states is a[1]*a[2]*...*a[n]. And each time the state shifts a lot.
It's a great time, but it's a viable approach.

Of course, we'll talk about this topic here, which means it's definitely not that complicated. Yes, there's a very magical conclusion to this game:

for a situation, when and only if A[1] xor a[2] xor ... xor a[n] = 0 o'clock, the situation is p.

The evidence for this conclusion is as follows:
1. All 0 states are P-situation, i.e. a[i]=0, then a[1] xor a[2] xor ... xor a[n] = 0.
2. From any one a[1] xor a[2] xor ... xor a[n] = k! = 0 The state can be moved to A[1] XOR a[2] xor ... xor a[n] = 0 state. Because of the particularity of the XOR calculation, we know that there must be a a[i] the highest bit is the same as 1 of the highest bit of K, then there is bound to be a[i] XOR K < a[i], so we can change the value of a[i] to A[i] ', make a[1] xor A[2] xor ... xor a[ I] ' xor ... xor a[n] = 0.
3. For any situation, if A[1] xor a[2] xor ... xor a[n] = 0, then there is no one move can make the new situation a[1] XOR A[2] xor ... xor a[n]! = 0. Because of the particularity of the XOR calculation, we can know that there must be an even number of 1 o'clock in this position 1 will be eliminated. Changing only one a[i] will, in any case, change the number of 1, leading to a[1] XOR a[2] xor ... xor a[n]! = 0.
The above three satisfies the transfer nature of the n,p situation in the ICG game, so the correctness of this conclusion is also proved.

The code is as follows

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace Std;
int a,n,p;
int main ()
{
cin>>n;
cin>>p;
n--;
while (n>0) {
cin>>a;
P^=a;
n--;
}
if (p==0) cout<< "B" <<endl;
else cout<< "A" <<endl;
return 0;
}

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