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  • hdu-1907 John

    题目链接:

    http://acm.hdu.edu.cn/showproblem.php?pid=1907

    题目类型:

    博弈

    题意描述:

    有N个装有若干个糖果的盒子,有两个人轮流取盒子中的糖果,每次只能从一个盒子中拿不少于1个糖果(可以一次性取完)直到所有糖果都取完,最后一次取糖果的人输,问john是胜是负。

    解题思路:

    该题是nim博弈的变形,先进行一次判断是否所有的盒子中全都是1个糖果,如果这样,结果已定,如果不是,则对每一堆糖果的个数进行异或运算,如果最终值为0,则John输,反之则John胜。

    题目:

    John

    Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
    Total Submission(s): 4816    Accepted Submission(s): 2780


    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
     
    # include <stdio.h>
    
    int game(int a[],int n)
    {
        int i,num,ret;
        ret=0;
        for(i=0;i<n;i++)
            if(a[i]!=1)
                ret=1;
        if(ret==0)
        {
            if(n%2==0)
                return 1;
            else
                return 0;
        }
        num=0;
        for(i=0;i<n;i++)
            num=num^a[i];
        if(num!=0)
            return 1;
        else
            return 0;
    }
    
    int main ()
    {
        int t,n,i,a[50];
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&n);
            for(i=0;i<n;i++)
                scanf("%d",&a[i]);
            if(game(a,n)==1)
                printf("John
    ");
            else
                printf("Brother
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/love-sherry/p/6965790.html
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