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 note 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.

博弈

 #include<stdio.h>
#include<string.h> int main(){
int T;
while(scanf("%d",&T)!=EOF){
while(T--){
int n;
scanf("%d",&n);
int i,num=,sum=;
for(i=;i<=n;i++){
int a;
scanf("%d",&a);
sum^=a;
if(a!=)num++;
}
if((num==&&sum==)||(sum!=&&num>))printf("John\n");
else printf("Brother\n");
}
}
return ;
}

hdu1907 John 博弈的更多相关文章

  1. HDU1907 John

    Description Little John is playing very funny game with his younger brother. There is one big box fi ...

  2. 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. ======== ...

  3. 尼姆博弈HDU1907

    HDU1907 http://acm.hdu.edu.cn/showproblem.php?pid=1907 两种情况1.当全是1时,要看堆数的奇偶性 2.判断是奇异局势还是非奇异局势 代码: #in ...

  4. John(博弈)

    Description Little John is playing very funny game with his  younger brother. There  is one big box ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. HDU1907(尼姆博弈)

    John Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submis ...

  9. POJ 3480 &amp; 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 ...

随机推荐

  1. Win10系列:JavaScript页面导航

    页面导航是在开发应用的过程中使用频率较高的技术,其中比较常用的导航方式有多页导航和页内导航,采用多页导航方式的应用程序包含一系列的页面,在一个页面中加入另一个页面的链接地址后,单击链接将跳转到指定页面 ...

  2. netty]--最通用TCP黏包解决方案

    netty]--最通用TCP黏包解决方案:LengthFieldBasedFrameDecoder和LengthFieldPrepender 2017年02月19日 15:02:11 惜暮 阅读数:1 ...

  3. 图的深度优先遍历(DFS)和广度优先遍历(BFS)

    body, table{font-family: 微软雅黑; font-size: 13.5pt} table{border-collapse: collapse; border: solid gra ...

  4. tomcat设置默认启动项

     Tomcat设置默认启动项目 Tomcat设置默认启动项目,顾名思义,就是让可以在浏览器的地址栏中输入ip:8080,就能访问到我们的项目.具体操作如下:     1.打开tomcat的安装根目 ...

  5. jdk8-stream的api

    1.stream流的概念 1.流的创建 //1. 创建 Stream @Test public void test1(){ //1. Collection 提供了两个方法 stream() 与 par ...

  6. SQL-7查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数t (group 与count)

    题目描述 查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数tCREATE TABLE `salaries` (`emp_no` int(11) NOT NULL,`salary` int ...

  7. Centos7安装ansible

    CentOS下部署Ansible自动化工具 1.确保机器上安装的是 Python 2.6 或者 Python 2.7 版本: python -V 2.查看yum仓库中是否存在ansible的rpm包 ...

  8. Docker(2):快速入门及常用命令

    什么是Docker? Docker 是世界领先的软件容器平台.开发人员利用 Docker 可以消除协作编码时“在我的机器上可正常工作”的问题.运维人员利用 Docker 可以在隔离容器中并行运行和管理 ...

  9. 牛客多校第四场 F Beautiful Garden

    链接:https://www.nowcoder.com/acm/contest/142/F来源:牛客网 题目描述 There's a beautiful garden whose size is n ...

  10. Unity3d代码及效率优化总结

    1.PC平台的话保持场景中显示的顶点数少于200K~3M,移动设备的话少于10W,一切取决于你的目标GPU与CPU. 2.如果你用U3D自带的SHADER,在表现不差的情况下选择Mobile或Unli ...