标签：acm 尼姆博弈 john hdu1907
John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2577 Accepted Submission(s): 1400
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
Sample Output
Source
近期在看博弈系列，搞完了巴什博弈、威佐夫博弈，接下来就是这个尼姆博弈。
这一系列博弈类型可以概括为：
巴什博弈：从一堆石子中拿石子，一次拿1~m个。
威佐夫博弈：从两堆石子中拿石子，方法①任选一堆石子拿k个石子（k≥1），方法②从两堆石子中拿相同数量的石子（当然所拿的数要≥1）
尼姆博弈：从三堆石子中拿石子，每次任选一堆拿任意数目（≥1）的石子。
这些当然是谁先那光，谁获胜。
尼姆博弈解法，和二进制有关。
反正我不知道他们怎么推得，只会用。o(╯□╰)o。。。
给出的数 用异或加起来，若等于0，则为奇异态（必胜态）。
这道题有些不一样，如果John吃的是某个盒子最后一颗，那就判定John为败。
所以，这道题分为两种情况讨论：
①若所有堆的数量都为1。则根据奇偶来判断谁胜。
②其他情况，将所有数据异或起来，判断是否为奇异态。
/**************************************
***************************************
* Author:Tree *
*From :http://blog.csdn.net/lttree *
* Title : John *
*Source: hdu 1907 *
* Hint : 尼姆博弈 *
***************************************
**************************************/
#include <iostream>
#include <algorithm>
using namespace std;
int arr[48];
int main()
{
int t,n,i,temp;
cin>>t;
while( t-- )
{
cin>>n;
for(i=0;i<n;++i)
cin>>arr[i];
sort(arr,arr+n);
// 如果全是1，按照奇偶判断谁获胜
if( arr[n-1]==1 )
{
if( n&1 ) cout<<"Brother"<<endl;
else cout<<"John"<<endl;
continue;
}
// 异或加起来
temp=arr[0]^arr[1];
for(i=2;i<n;++i)
temp^=arr[i];
if( temp==0 ) cout<<"Brother"<<endl;
else cout<<"John"<<endl;
}
return 0;
}
ACM-尼姆博弈之John——hdu1907,布布扣,bubuko.com
ACM-尼姆博弈之John——hdu1907
标签：acm 尼姆博弈 john hdu1907
原文地址：http://blog.csdn.net/lttree/article/details/24874819
