zoj 2507 Let's play a game

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1507

Elves from the Lothvain forest have created a very interesting game. The rules are very simple:

There are two players.

• In the beginning there are n groups of stones. The i-th group contains a_i stones.

• A player has to take a positive number of stones from exactly one of the groups in his turn.

• A player who can’t make a move (each group is empty) wins.

Very soon they have learnt how to always make the best possible move. Your task is to write a program that computes which of the players has a winning strategy for a given situation.

INPUT

The first number appearing in the input is number of datasets t (5<=t<=50). Each dataset is given in following format. The first row contains a single number n (1<=n<=10000). The next row contains exactly n numbers: a_1, a_2,…, a_n (0<=a_i<=1000000000). There are no empty lines between the datasets. You may assume that at least one group of stones is not empty.

OUTPUT

For each of datasets your program should write:

• 1 if the first player has a winning strategy

• 2 otherwise

in a separate row.

SAMPLE INPUT
2
3
1 1 1
5
1 0 1 1 1

SAMPLE OUTPUT
2
1

题意：NIM游戏的变种，胜利状态变为不能取的取胜。

先手必胜当且仅当：

1、若所有堆的石子最大数量为1，且数量为1的堆数为偶数

2、若有的堆数量超过1，SG为非0
证明：情况一、很容易想到。

情况二、先分析只有一堆石子的数量大于1，那么这个人可以控制后手的1的数量，因此是必胜态（SG必为非0）。

其次，如果有大于1堆的石子的数量大于2，那么如果SG为非0，那么先手只要将SG变为0，后手无论进行什么操作，SG都会变成非0（异或的性质），这样进行下去，终会达到必胜态（只有一堆石子的最大数量大于1）。

#include<iostream>
#include <algorithm>
#include <cstdio>
#include <string.h>
using namespace std;
long long  a[10005];
int main()
{
#ifndef ONLINE_JUDGE
    //freopen("1.txt", "r", stdin);
#endif
    int n, i, j, t;
    long long ans;
    cin >> t;
    bool flag;
    while(t--)
    {
        flag = false;
        cin >> n;
        ans = 0;
        for (i = 0; i < n; i++)
        {
            cin >> a[i];
            ans ^= a[i];
            if (a[i] > 1)
            {
                flag = true;
            }
        }
        if (flag)
        {
            if (ans)
            {
                cout << "1\n";
            }
            else
            {
                cout << "2\n";
            }
        }
        else
        {
            if (ans)
            {
                cout << "2\n";
            }
            else
            {
                cout << "1\n";
            }
        }
    }

    return 0;
}