POJ 1704：Georgia and Bob

Description

Georgia and Bob decide to play a self-invented game. They draw a row of grids on paper, number the grids from left to right by 1, 2, 3, …, and place N chessmen on different grids, as shown in the following figure for example: 
Georgia and Bob move the chessmen in turn. Every time a player will choose a chessman, and move it to the left without going over any other chessmen or across the left edge. The player can freely choose number of steps the chessman moves, with the constraint that the chessman must be moved at least ONE step and one grid can at most contains ONE single chessman. The player who cannot make a move loses the game.

Georgia always plays first since “Lady first”. Suppose that Georgia and Bob both do their best in the game, i.e., if one of them knows a way to win the game, he or she will be able to carry it out.

Given the initial positions of the n chessmen, can you predict who will finally win the game?

Input

Output

Sample Input

2
3
1 2 3
8
1 5 6 7 9 12 14 17

Sample Output

Bob will win
Georgia will win

给一个1*M的棋盘，上面有N颗棋子，每次只能向左移动棋子，并且至少移动一步，两人轮流操作，谁不能移动谁就输了。

很明显的博弈问题，我们可以把整个棋盘中相邻棋子之间的距离计算出来，人为操作可能会使这个距离增大，也可能会让它减小，但是，每两次操作中，不论如何移动棋子，我们都可以假设这个距离一定会减小，这样的话，这个距离就相当于石子的数量，两人轮流取走石子，看谁取走最后的石子啦！问题就这样转换成了Nim博弈，但是如果棋盘上的棋子数量是奇数应该怎么办呢？我们可以添加一个棋子，在棋盘的最左端，也就是它的位置是0，每堆石子的数量可认为是相邻每两个棋子的距离差，然后计算异或，判断数值之后得出结果。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        int a[n+10],s=0;
        for(int i=0; i<n; i++)
            cin>>a[i];
        if(n&1)a[n++]=0;
        sort(a,a+n);
        for(int i=1; i<n; i+=2)
            s^=a[i]-a[i-1]-1;
        if(s)cout<<"Georgia will win"<<endl;
        else cout<<"Bob will win"<<endl;
    }
    return 0;
}