寒光博客

[POJ]1704:Georgia and Bob
Staircase Nim Problem Description Georgia and Bob decide...
扫描右侧二维码阅读全文
23
2019/07

[POJ]1704:Georgia and Bob

Staircase Nim

Problem 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:
map
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

分析


坑点 就是 数据需要排序 ,,,哭了

代码

package POJ;

import java.util.Arrays;
import java.util.Scanner;

public class _1704_GeorgiaAndBob {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int k=cin.nextInt();
        for (int i = 0; i < k; i++) {//几次数据测试
            int ans=0;
            int n = cin.nextInt();
            int[] num = new int[n];
            for (int j = 0; j < n; j++) {//n个 输入
                num[j] = cin.nextInt();
            }
            Arrays.sort(num);//坑点  要排序
            if (n % 2 == 0) {//偶数 12 34 6 9
                for (int j = 1; j < n; j+=2) {
                    ans^=(num[j]-num[j-1]-1);
                }
            }else{
                for (int j = 0; j <n ; j+=2) {//2 4 5
                    ans^=(j==0)?(num[j]-1):(num[j]-num[j-1]-1);
                }
            }
            if(ans==0){//先手输了
                System.out.println("Bob will win");
            }else
                System.out.println("Georgia will win");
        }
    }
}

题目地址:
POJ1704:https://vjudge.net/problem/POJ-1704
本文作者:Author:     文章标题:[POJ]1704:Georgia and Bob
本文地址:https://dxoca.cn/Algorithm/182.html       百度未收录
版权说明:若无注明,本文皆为“Dxoca's blog (寒光博客)”原创,转载请保留文章出处。
Last modification:July 31st, 2019 at 12:16 pm
如果觉得我的文章对你有用,请随意赞赏

Leave a Comment