John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2179 Accepted Submission(s): 1175
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.
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
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
先者必败条件： 奇异局势&& 充裕堆数目>=2 或者 非奇异局势&& 孤单堆数目==n
/***************************
# 2013-8-23 17:04:27
# Time: MS Memory: KB
# Author: zyh
# Status: Accepted
***************************/
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int main()
{
int n,m,t,sum,cnt1,cnt2,flag,i,a[48];
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(sum=cnt1=cnt2=0,i=0;i<n;i++){
scanf("%d",&a[i]);
if(a[i]>1) cnt1++; //充裕堆数目
else cnt2++; //孤单堆数目
sum ^= a[i];
}
if((sum==0&&cnt1>=2) || (sum && cnt2==n) )printf("Brother\n");
else printf("John\n");
}
return 0;
}