标签：-- pair amp fine map 博弈 limit ima led
John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 6000 Accepted Submission(s): 3486
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
C/C++：
1 #include <map>
2 #include <queue>
3 #include <cmath>
4 #include <vector>
5 #include <string>
6 #include <cstdio>
7 #include <cstring>
8 #include <climits>
9 #include <iostream>
10 #include <algorithm>
11 #define INF 0x3f3f3f3f
12 #define LL long long
13 #define wzf ((1 + sqrt(5.0)) / 2.0)
14 using namespace std;
15 const int MAX = 100;
16
17 int n, t, num[MAX], temp;
18
19 int main()
20 {
21 scanf("%d", &t);
22 while (t --)
23 {
24 scanf("%d", &n);
25 for (int i = 0; i < n; ++ i)
26 scanf("%d", &num[i]);
27 sort(num, num + n);
28 if (num[n - 1] == 1)
29 {
30 if (n & 1) printf("Brother\n");
31 else printf("John\n");
32 continue;
33 }
34 temp = num[0] ^ num[1];
35 for (int i = 2; i < n; ++ i)
36 temp ^= num[i];
37 if (temp == 0)
38 printf("Brother\n");
39 else
40 printf("John\n");
41 }
42 return 0;
43 }
hdu 1907 John (尼姆博弈)
标签：-- pair amp fine map 博弈 limit ima led
原文地址：https://www.cnblogs.com/GetcharZp/p/9557856.html
