• hdu 1907 John (尼姆博弈)


    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
    ");
    31             else printf("John
    ");
    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
    ");
    39         else
    40             printf("John
    ");
    41     }
    42     return 0;
    43 }
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  • 原文地址:https://www.cnblogs.com/GetcharZp/p/9557856.html
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