hdu1907 John （尼姆博弈）

John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 4183 Accepted Submission(s): 2385

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：n堆石子，每次可以从一堆石子中选取若干，最少1个，谁最后拿完谁输.

思路：尼姆博弈，只不过判断一下全部为1的情况.

/* ***********************************************
Author       : AnICoo1
Created Time : 2016-07-25-19.51 Monday
File Name    : D:\MyCode\No.2\main.cpp
LANGUAGE     : C++
Problem      : hdu 1907
Copyright  2016 clh All Rights Reserved
************************************************ */
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head

int main()
{
int t;cin>>t;
while(t--)
{
int n;cin>>n;
int ans=0;int x=0,y=0;
for(int i=0;i<n;i++)
{
int a;cin>>a;
ans^=a;
if(a==1) x++;
}
if(x==n)
{
if(x%2) printf("Brother\n");
else printf("John\n");
continue;
}
if(ans)
printf("John\n");
else
printf("Brother\n");
}
return 0;
}