HDU5795-A Simple Nim（SG函数）

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本文介绍了一种名为ASimpleNim的游戏，玩家通过取石子进行对决。文章提供了问题描述、输入输出样例及解题思路。核心在于发现特定规律以确定胜者。

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A Simple Nim

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1464 Accepted Submission(s): 790

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

Sample Output

  
   Second player wins.
First player wins.

Author

UESTC

Source

2016 Multi-University Training Contest 6

Recommend

wange2014

题意：取石子游戏，n堆石头，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。
解题思路：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)。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int sg[100];
bool vis[100];

int main()
{
	//打表
	/*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 j = 1; j < i; j++)
		{
			for (int k = 1; k < j; k++)
			{
				int u = i - j - k;
				if (u > 0)
				{
					int tmp = sg[j] ^ sg[k] ^ sg[u];
					vis[tmp] = 1;
				}
				else break;
			}
		}
		for (int j = 0;; j++)
			if (!vis[j])
			{
				sg[i] = j;
				printf("sg[%d]: %d\n", i, x);
				break;
			}
	}*/
	int t, n;
	scanf("%d", &t);
	while (t--)
	{
		int ans = 0, a;
		scanf("%d", &n);
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &a);
			if (a % 8 == 7) ans ^= (a + 1);
			else if (a % 8 == 0) ans ^= (a - 1);
			else ans ^= a;
		}
		if (ans) printf("First player wins.\n");
		else printf("Second player wins.\n");
	}
	return 0;
}