取石子游戏(找规律+sg函数+博弈)HDU 5795 A Simple Nim

题面：

A Simple Nim Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 300    Accepted Submission(s): 211


Problem Description Two players take turns picking candies from n heaps,the player who picks the last one will win the game.On each turn they can pick any number of candies which come from the same heap(picking no candy is not allowed).To make the game more interesting,players can separate one heap into three smaller heaps(no empty heaps)instead of the picking operation.Please find out which player will win the game if each of them never make mistakes.  
Input Intput 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≤106,1≤s[i]≤109)  
Output For each test case,output a line whick contains either"First player wins."or"Second player wins".  
Sample Input

2 2 4 4 3 1 2 4  
Sample Output

Second player wins. First player wins.  
Author UESTC  
Source 2016 Multi-University Training Contest 6  
题意：
    取石子游戏，有两种操作方式，一、在一堆中取任意颗石子，（大于0）。二、将一堆分成三堆，每堆数量大于0。取到最后一块石子的人获得胜利。

解题：
    先小数据打表求sg值，可以发现sg值的规律。当i%8==7时，其sg值为i+1，当i%8==0时，其sg值为i-1(sg[0]=0)。一个状态的sg值，是其后继状态sg值中未出现过最小整数，三堆的sg值是三堆石子数量sg值的异或。根据sg值的规律，可以求解问题，官方题解说是用数学归纳法证明。

代码：

#include#include#include#define LL long long
#define mod 1000000007
#define sz 100005
using namespace std;
int sg[sz];
bool vis[sz];
int main()
{
	//打表程序
	/*int tmp;
    sg[0]=0;
	for(int i=1;i<=50;i++)
	{
		memset(vis,0,sizeof(vis));
		for(int j=0;j<i;j++)
			vis[sg[j]]=1;
		for(int k=1;k<i;k++)
		{
			for(int m=1;m0)
                {
					tmp=sg[k]^sg[m]^sg[u];
					vis[tmp]=1;
				}
				else
					break;
			}
		}
		for(int x=0;;x++)
			if(!vis[x])
			{
				sg[i]=x;
				printf("sg[%d]: %dn",i,x);
				break;
			}
	}*/
	int t,n,tmp,s;
	scanf("%d",&t);
    while(t--)
	{
		s=0;
		scanf("%d",&n);
        while(n--)
		{
			scanf("%d",&tmp);
			if(tmp%8==7)
				s^=(tmp+1);
			else if(tmp%8==0)
				s^=(tmp-1);
			else
				s^=tmp;
		}
		if(s)
			printf("First player wins.n");
		else
			printf("Second player wins.n");
	}
	return 0;
}