HDU 1907:John(尼姆博弈变形)
John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 6017 Accepted Submission(s): 3499
Problem 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 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.
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.
Constraints:
1 <= T <= 474,
1 <= N <= 47,
1 <= Ai <= 4747
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.
Sample Input
2
3
3 5 1
1
1
Sample Output
John
Brother
题意
两个人取n堆石子,每个人至少去一个,最多把一堆石子取完,取到最后一个石子的人失败
思路
先手胜的情况:
- n堆石子全部都只有一个石子,且n堆石子的异或值为0
- n堆石子不全是一个石子,且异或值不为0
证明:
- 若所有堆石子数都为1且SG值为0,则共有偶数堆石子,故先手胜。
- 只有一堆石子数大于1时,我们总可以对该堆石子操作,使操作后石子堆数为奇数且所有堆得石子数均为1
- 有超过一堆石子数大于1时,先手将SG值变为0即可,且总还存在某堆石子数大于1
思路来自:http://hzwer.com/1950.html
AC代码
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ull unsigned long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x7f7f7f7f
#define lson o<<1
#define rson o<<1|1
const double E=exp(1);
const int maxn=1e6+10;
const int mod=1e9+7;
using namespace std;
int main(int argc, char const *argv[])
{
ios::sync_with_stdio(false);
int t;
int n;
int x;
cin>>t;
while(t--)
{
cin>>n;
int sum=0;
int res=0;
while(n--)
{
cin>>x;
sum^=x;
if(x>1)
res++;
}
if(!res)
{
if(!sum)
cout<<"John"<<endl;
else
cout<<"Brother"<<endl;
}
else
{
if(!sum)
cout<<"Brother"<<endl;
else
cout<<"John"<<endl;
}
}
return 0;
}
HDU 1907:John(尼姆博弈变形)的更多相关文章
- hdu 1907 John (尼姆博弈)
John Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submis ...
- hdu 1907 (尼姆博弈)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1907 Problem Description Little John is playing very ...
- 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 1849 (尼姆博弈)
http://acm.hdu.edu.cn/showproblem.php? pid=1849 简单的尼姆博弈: 代码例如以下: #include <iostream> #include ...
- HDU 1907 John (Nim博弈)
John Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submis ...
- John 尼姆博弈
John Little John is playing very funny game with his younger brother. There is one big box filled wi ...
- 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 ...
- hdu 1907 尼姆博弈
John Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submis ...
- HDU.1850 being a good boy in spring festival (博弈论 尼姆博弈)
HDU.1850 Being a Good Boy in Spring Festival (博弈论 尼姆博弈) 题意分析 简单的nim 博弈 博弈论快速入门 代码总览 #include <bit ...
随机推荐
- vs 编译库文件 Qt编译库文件
QT 库能不能用 需要关注是minGW 还是MSVC编译的 Qt MinGW与MSVC对比 转:https://blog.csdn.net/u013185164/article/details/48 ...
- Ubuntu中的在文件中查找和替换命令
分类: 9.Linux技巧2009-09-29 13:40 1429人阅读 评论(0) 收藏 举报 ubuntujdbc 1.查找 find /home/guo/bin -name /*.txt | ...
- Fiddler系列教程1:初识Http协议抓包工具
1. Fiddler简介 Fiddler是用一款使用C#编写的http协议调试代理工具.它支持众多的http调试任务,能够记录并检查所有你的电脑和互联网之间的http通讯,可以设置断点,查看所有的“进 ...
- mybatis if标签判断字符串相等
mybatis 映射文件中,if标签判断字符串相等,两种方式: 因为mybatis映射文件,是使用的ognl表达式,所以在判断字符串sex变量是否是字符串Y的时候, <if test=" ...
- nginx+tomcat集群
参考: 简单:http://blog.csdn.net/wang379275614/article/details/47778201 详细:http://www.jb51.net/article/77 ...
- sqlserver指定排序字段
在sqlserver中可以指定排序的字段,需要将哪个字段值排在最前面或最后面,都是可以的.见如下代码: SELECT * FROM public_comment order by case [User ...
- (C/C++学习笔记) 一. 基础知识
一. 基础知识 ● 程序和C/C++ 程序: 根据Wirth (1976), Algorithms + Data Structures = Programs. Whence C: 1972, Denn ...
- 在ros功能包CMakeLists.txt中获取所在功能包的路径 便于添加第三方库的相对路径
在 ros 功能包中要使用第三方的动态库,将其放在系统默认库路径和使用绝对路径均不可取,这样的话可移植性较差,将该功能包移到其它电脑时要重新配置依赖库的路径,太麻烦了. 于是找到下面这个方法,解决了R ...
- python笔记3-输出输入、字符串格式化
输入.输出 python怎么来接收用户输入呢,使用input函数,python2中使用raw_input,接收的是一个字符串,输出呢,第一个程序已经写的使用print,代码入下: 1 2 name=i ...
- python学习二三事儿(转,整)
Python 标识符 在 Python 里,标识符由字母.数字.下划线组成. 在 Python 中,所有标识符可以包括英文.数字以及下划线(_),但不能以数字开头. Python 中的标识符是区分大小 ...