John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 5793 Accepted Submission(s): 3358
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
3
3 5 1
1
1
Sample Output
John
Brother
Brother
Source
对于所有不同颜色M&M的个数都为1的情况,显然:
if(sg==0) win -->状态A
if(sg!=0) lose -->状态B
对于对立情况，
当数量不为1的种类只有一种的时候(状态C)，显然先手总可以将其转化到状态B这个败态，所以C是胜态。
当数量不为1的种类大于1的时候,分为sg=0(状态D)和sg!=0(状态E)两种情况，D只能转化到E,E能转化到D或者
C,由于C是败态，所以E尽可能向D转化，但随着石子的不断减少导致E只能转化到C，所以E是败态，D是胜态。
#include<bits/stdc++.h>
using namespace std;
int main(){
int t,n,m;
cin>>t;
while(t--){
int a,sg=,tot=;
cin>>n;
for(int i=;i<=n;++i){
cin>>a;
sg^=a;
if(a==) tot++;
}
if( (tot==n&&sg==) || (tot!=n&&sg!=)) puts("John");
else puts("Brother");
}
return ;
}