POJ

Georgia and Bob

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 11093 Accepted: 3661

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

231 2 381 5 6 7 9 12 14 17

Sample Output

Bob will winGeorgia will win

题意： 给你n个石子的位置 对于每一个石子如果他的前边有空位那么他就可以向前

移动，两个人轮流移动直到一个人不能移动为止（为输） geo先移动  bob 后移动

思路： 很经典的阶梯博弈。（nim博弈的变形）

推荐一篇博客  阶梯博弈

代码： 

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define N 10005using namespace std;int num[N];int a[N];int n;int main(){    int cas;    cin>>cas;    while(cas--)    {        cin>>n;        for(int i=1;i<=n;i++) cin>>a[i];        sort(a+1,a+n+1);        for(int i=n;i>=1;i--) num[i]=a[i]-a[i-1]-1;        int fin=0;        for(int i=n;i>=1;i-=2) fin^=num[i];        if(fin) printf("Georgia will win\n");        else printf("Bob will win\n");    }    return 0;}