John
John |
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) |
Total Submission(s): 60 Accepted Submission(s): 41 |
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.
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Sample Input
2 3 3 5 1 1 1 |
Sample Output
John Brother |
Source
Southeastern Europe 2007
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lcy
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/* 题意:有一个盒子,里面装着不同颜色的糖,没人每次只能吃相同颜色的糖。并且最少吃一个,谁吃到最后一个糖就会输 就输。 初步思路:就是裸的尼姆博弈,石头堆换成相同颜色的糖 #错误:少了特判,如果都是1的话,就不需要看异或了。 */ #include<bits/stdc++.h> using namespace std; int t,n,a[50]; int main(){ // freopen("in.txt","r",stdin); scanf("%d",&t); int res=0; while(t--){ scanf("%d",&n); res=0; int flag=0; for(int i=0;i<n;i++){ scanf("%d",&a[i]); res^=a[i]; if(a[i]>1) flag=1; } if(flag){ printf(res?"John\n":"Brother\n"); }else{ printf(n&1?"Brother\n":"John\n"); } } return 0; }
我每天都在努力,只是想证明我是认真的活着.