John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 4704 Accepted Submission(s): 2720
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
Source
1 #include
2 using namespace std;
3
4 int main()
5 {
6 int t;
7 int n;
8 int a;
9 int i;
10 int sum, ones;
11
12 scanf("%d", &t);
13
14 while (t--) {
15 scanf("%d", &n);
16 sum = 0, ones = 0;
17 for (i = 0; i < n; ++i) {
18 scanf("%d", &a);
19 sum ^= a;
20 if (a == 1) {
21 ++ones;
22 }
23 }
24
25 //win : t0, s1, s2
26 //lose : s0, t2
27 if (sum == 0) {//t
28 if (n - ones == 0) {//t0
29 printf("John\n");
30 } else {//t2
31 printf("Brother\n");
32 }
33 } else {//s
34 if (n - ones == 0) {//s0
35 printf("Brother\n");
36 } else {//s1, s2
37 printf("John\n");
38 }
39 }
40 }
41
42 return 0;
43 }
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