Nim or not Nim?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Nim is a two-player mathematic game of strategy in which players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come
from the same heap.

Nim is usually played as a misere game, in which the player to take the last object loses. Nim can also be played as a normal play game, which means that the person who makes the last move (i.e., who takes the last object) wins. This is called normal play because
most games follow this convention, even though Nim usually does not.

Alice and Bob is tired of playing Nim under the standard rule, so they make a difference by also allowing the player to separate one of the heaps into two smaller ones. That is, each turn the player may either remove any number of objects from a heap or separate
a heap into two smaller ones, and the one who takes the last object wins.

Input

Input contains multiple test cases. The first line is an integer 1 ≤ T ≤ 100, the number of test cases. Each case begins with an integer N, indicating the number of the heaps, the next line contains N integers s[0], s[1], ...., s[N-1],
representing heaps with s[0], s[1], ..., s[N-1] objects respectively.(1 ≤ N ≤ 10^6, 1 ≤ S[i] ≤ 2^31 - 1)

Output

For each test case, output a line which contains either "Alice" or "Bob", which is the winner of this game. Alice will play first. You may asume they never make mistakes.

Sample Input

2
3
2 2 3
2
3 3

Sample Output

Alice
Bob

由于数据范围比较大，如果直接求SG函数的话，会MLE，，，但是我们可以先打表求出部分SG函数，然后找规律，

SG打表代码：

#include <stdio.h>
#include <string.h>
#define MAX 50
int sg[MAX] ;
int getSG()
{
	bool visited[100] ;
	for(int i = 1 ; i < MAX ; ++i)
	{
		memset(visited,false,sizeof(visited)) ;
		for(int j = 1 ; j <= i ; ++j)
		{
			visited[sg[i-j]] = true ;
		}
		for(int j = 1 ; j < i ; ++j)
		{
			visited[sg[i-j]^sg[j]] = true ;
		}
		for(int j = 1 ; j < MAX ; ++j)
		{
			if(!visited[j])
			{
				sg[i] = j ;
				printf("%d %d\n",i,sg[i]) ;
				break ;
			}
		}
	}
}

int main()
{
	getSG() ;
	return 0 ;
}

打表结果：

当x%4=1or2时，sg[x] = x,当x%4==3时，sg[x]=x+1,当x%4==0时，sg[x] = x-1；

所以这道题的代码为：

#include <stdio.h>
#include <limits.h>
int main()
{
	int t ;
	scanf("%d",&t) ;
	while(t--)
	{
		int n ,ans = 0;
		scanf("%d",&n);
		for(int i = 0 ; i < n ; ++i)
		{
			int temp ;
			scanf("%d",&temp);
			if(temp%4==1||temp%4==2)
			{
				ans ^= temp ;
			}
			else if(temp%4==3)
			{
				ans ^= temp+1;
			}
			else if(temp%4==0)
			{
				ans ^= temp-1 ;
			}
		}
		if(ans)
		{
			puts("Alice") ;
		}
		else
		{
			puts("Bob") ;
		}
	}
	return 0 ;
}