POJ1704 Georgia and Bob(Nim博弈变形)

Georgia and Bob

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 14312		Accepted: 4840

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

Source

POJ Monthly--2004.07.18

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题解：

我们把相邻两给位置组成一对，如果N为奇数的话，就让第一个数和0组成一对，然后考虑每一对，无论前面的位置移动多少，后面一个点总能移动相同的距离使得两个点对之间的距离保持不变，所以我们就可以将每一对之间的距离拿出来，就变成了了这(N+1)/2个数之间的NIM博弈了，异或即可，很巧妙的一道题。

参考代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn=;
int T,n,a[maxn],ans;

int main()
{
    scanf("%d",&T);
    while(T--)////
    {
        scanf("%d",&n);
        for(int i=;i<=n;++i) scanf("%d",&a[i]);

        sort(a+,a+n+);
        ans=; a[]=;
        for(int i=n;i>;i-=)
            ans^=(a[i]-a[i-]-);
        if(ans) puts("Georgia will win");
        else puts("Bob will win");
    }

    return ;
}