POJ1704 Georgia and Bob (阶梯博弈)

Posted 2020-07-31 Ritchie丶

Georgia and Bob

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

Submit Status

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

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case contains two lines. The first line consists of one integer N (1 <= N <= 1000), indicating the number of chessmen. The second line contains N different integers P1, P2 ... Pn (1 <= Pi <= 10000), which are the initial positions of the n chessmen.

Output

For each test case, prints a single line, "Georgia will win", if Georgia will win the game; "Bob will win", if Bob will win the game; otherwise ‘Not sure‘.

Sample Input

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

Sample Output

Bob will win
Georgia will win
首先认识一下什么是阶梯博弈。注意这个网址中所说的“2^3^4=5 不为零所以先手必败”是错的
异或不为0应该是先手必胜，说反了。
http://blog.csdn.net/kk303/article/details/6692506

题目大意：

每个测试点最多有T（1 <= T <= 20）个测试数据。如图所示，Georgia和Bob在玩一种自创的游戏。一个无限长的棋盘上有N个旗子（1 <= N <= 1000），第i个棋子的位置可以用Pi表示（1 <= Pi <= 10000）。现在Georgia先走。每个人每一次可以把一枚棋子向左移动任意个格子，但是不能超越其他棋子，也不能和其他棋子处在同一个格子里。如果轮到某一个人的时候Ta再也不能移动棋子了，就判负。现在每个测试数据给定一种情况，如果Georgia会赢，输出“Georgia will win”，如果Bob会赢，输出“Bob will win”，如果不确定，输出“Not sure”。两个人都知道获胜策略是什么，也会想方设法取得胜利。
思路：

以第二个样例为例：

1 5 6 7 9 12 14 17

第一个棋子不能向左移动了。第二个棋子可以向左移动3个格子。第三个棋子也不能移动了，以此类推，可以得到这样一个数列：

0 3 0 0 1 2 1 2，第n个数字代表第n个棋子可以移动的步数。

考虑一下把第二个棋子向左移动一格的情况，原数列变为：

0 2 1 0 1 2 1 2

这不就是把“第二堆”石子移了一个到右边的“第三堆”石子么？由此可以给出等价的游戏新定义：

给定N堆石子，每堆里面的石子个数都是非负的。每次可以把第i堆中的任意颗石子移动到第i + 1堆中（1 <= i < N），或者第N堆的石子扔掉任意颗。如果某人不能继续操作则判负。

注意最右面的棋子不是一直不动的，因为要保证异或值为0，所以最右面的棋子也要动。

把这个问题看做是每一个石子相对于前一个石子运动。

#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        //注意在本题中若循环跑的是1到n 在运算过程中一定不能视a[0]默认为0 新定义的数组a[0]不一定是0
        int n,a[1005],ans=0;
        cin>>n;
        for(int i=0;i<n;i++)
        cin>>a[i];
        sort(a,a+n);
        for(int i=n-1;i>=1;i--)
        a[i]=a[i]-a[i-1]-1; 
        a[0]-=1;
        for(int i=n-1;i>=0;i-=2)
        ans^=a[i];
        if(ans)
        cout<<"Georgia will win"<<endl;
        else
        cout<<"Bob will win"<<endl;
    }
    return 0;
}