John(博弈)
Description
Little John is playing very funny game with his younger brother. There is one big box filled
with M&Ms of different colors. At first John has to eat several M&Ms of the same color. Then his
opponent has to make a turn. And so on. Please no te that each player has to eat at least one
M&M during his turn. If John (or his brother) will eat the last M&M from the box he will be
considered as a looser and he will have to buy a new candy box.
Both of players are using optimal game strategy. John starts first always. You will be given
information about M&Ms and your task is to determine a winner of such a beautiful game.
Input
The first line of input will contain a single integer T – the number of test cases. Next T pairs of
lines will describe tests in a following format. The first line of each test will contain an integer N –
the amount of different M&M colors in a box. Next line will contain N integers Ai, separated by
spaces – amount of M&Ms of i-th color.
Output
Output T lines each of them containing information about game winner. Print “John” if John
will win the game or “Brother” in other case.
Constrains:
1 <= T <= 474,
1 <= N <= 47,
1 <= Ai <= 4747
Sample Input
Sample Output
分析:博弈题,最后拿者输!
代码:
#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
int main()
{
//freopen("a.in","r",stdin);
//freopen("aa.out","w",stdout);
int t,n,i,in,cnt,ans;
while(scanf("%d",&t)!=EOF)
{
while(t--)
{
scanf("%d",&n);
cnt=0;ans=0;
for(i=0;i<n;i++)
{
scanf("%d",&in);
if(in==1)
cnt++;
ans^=in;
}
if(cnt==n)
{
printf("%s\n",(cnt%2!=1)?"John":"Brother");
continue;
}
if(ans!=0)
printf("John\n");
else
printf("Brother\n");
}
}
return 0;
}
John(博弈)的更多相关文章
- hdu1907 John 博弈
Little John is playing very funny game with his younger brother. There is one big box filled with M& ...
- POJ 3480 John [博弈之Nim 与 Anti-Nim]
Nim游戏:有n堆石子,每堆个数不一,两人依次捡石子,每次只能从一堆中至少捡一个.捡走最后一个石子胜. 先手胜负:将所有堆的石子数进行异或(xor),最后值为0则先手输,否则先手胜. ======== ...
- hdu 1907 John&& hdu 2509 Be the Winner(基础nim博弈)
Problem Description Little John is playing very funny game with his younger brother. There is one bi ...
- HDU - 1907 John 反Nimm博弈
思路: 注意与Nimm博弈的区别,谁拿完谁输! 先手必胜的条件: 1. 每一个小游戏都只剩一个石子了,且SG = 0. 2. 至少有一堆石子数大于1,且SG不等于0 证明:1. 你和对手都只有一种选 ...
- POJ 3480 John(SJ定理博弈)题解
题意:n堆石头,拿走最后一块的输 思路:SJ定理:先手必胜当且仅当:(1)游戏的SG函数不为0且游戏中某个单一游戏的SG函数大于1:(2)游戏的SG函数为0且游戏中没有单一游戏的SG函数大于1. 参考 ...
- POJ 3480 & HDU 1907 John(尼姆博弈变形)
题目链接: PKU:http://poj.org/problem? id=3480 HDU:http://acm.hdu.edu.cn/showproblem.php? pid=1907 Descri ...
- HDU 1907 John(博弈)
题目 参考了博客:http://blog.csdn.net/akof1314/article/details/4447709 //0 1 -2 //1 1 -1 //0 2 -1 //1 2 -1 / ...
- bzoj1022: [SHOI2008]小约翰的游戏John(博弈SG-nim游戏)
1022: [SHOI2008]小约翰的游戏John 题目:传送门 题目大意: 一道反nim游戏,即给出n堆石子,每次可以取完任意一堆或一堆中的若干个(至少取1),最后一个取的LOSE 题解: 一道 ...
- HDU 1907 John nim博弈变形
John Problem Description Little John is playing very funny game with his younger brother. There is ...
随机推荐
- 使用SeaJS实现模块化JavaScript开发(新)
本文转自张洋,因为SeaJS更新版本很快,所以原文中很多地方不太适用,在这里发布一个更新版. 前言 SeaJS是一个遵循CommonJS规范的JavaScript模块加载框架,可以实现JavaSc ...
- Sublime Text 3 若干问题解决办法
1.在高分屏下中文文件夹名显示异常问题解决办法 新买了个2K的屏,有些中文文件夹名全部变成了“口口”. 在“preferences” - "设置-用户" 添加 "dpi_ ...
- ajax跨域调用
http://redsky008.iteye.com/blog/1754328 http://www.cnblogs.com/mahatmasmile/archive/2013/03/29/29895 ...
- bzoj 3153: Sone1 Toptree
3153: Sone1 Time Limit: 40 Sec Memory Limit: 256 MBSubmit: 511 Solved: 202[Submit][Status][Discuss ...
- [原博客] POJ 2425 A Chess Game
题目链接题意:给定一个有向无环图(DAG),上面放有一些旗子,旗子可以重合,两个人轮流操作,每次可以把一个旗子从一个位置移动到相邻的位置,无法移动时输,询问先手是否必胜. 这道题可以把每个旗子看作单独 ...
- 【技术贴】Maven打包文件增加时间后缀
构建war包,或者jar包的,时候,maven会自动增加一个版本号和时间放在jar包后面比如poi-3.9-20131115.jar这样子,但是我自己打war包,总是给我生成一个快照的后缀report ...
- 单片机系统与标准PC键盘的接口模块设计
转自单片机系统与标准PC键盘的接口模块设计 概述 在单片机系统中,当输入按键较多时,在硬件设计和软件编程之间总存在着矛盾.对于不同的单片机系统需要进行专用的键盘硬件设计和编程调试,通用性差,使 ...
- Setup SQL Server 2008 Maintenance Plan Email Notifications
一条龙作完,如何设置EXCHANGE的操作员邮件通知.. ~~~~ http://808techblog.com/2009/07/setup-sql-server-2008-maintena.html ...
- 【Uvalive 5834】 Genghis Khan the Conqueror (生成树,最优替代边)
[题意] 一个N个点的无向图,先生成一棵最小生成树,然后给你Q次询问,每次询问都是x,y,z的形式, 表示的意思是在原图中将x,y之间的边增大(一定是变大的)到z时,此时最小生成数的值是多少.最后求Q ...
- 【Xamarin挖墙脚系列:Xamarin的核心】
原文:[Xamarin挖墙脚系列:Xamarin的核心] Xamarin 包含两个商业产品 :Xamarin.IOS, Xamarin.Android.他们都是通过开源的基于.Net的Mono项目构建 ...