Question

2个角硬币游戏接收一个数组。游戏的目标是最大程度地累积点数（数组中元素的值）。您只能从数组的两个角取点。

游戏有两个条件：

1）第一个玩家（同伴）永远不会输（他会赢或平局），也不一定会选择最大优势。

2）第二个玩家（塔玛拉）将始终在最高角处得分。

我的输出：

amir took 16
tamara took 23
amira took 30
tamara took 15
amir took 19
tamara took 21
amir took 14
tamara took 13
Final Score:
amir total 79
tamara total 72

预期输出：

Amir took 16
Tamara took 23
Amir took 30
Tamara took 15
Amir took 19
Tamara took 21
Amir took 13
Tamara took 14
Final Score:
Amir total 78
Tamara total 73

问题：

-amir将选择16，因此tamara必须选择23，因此在下一回合中，他将选择30（因为tamara始终选择最大的角）。

-amir会在最后一轮选择13，而不是14，因为他已经赢了，所以他不在乎积分/硬币价值。

阿米尔之所以选择13而没有选择14的原因： 1.因为塔玛拉总是会选择最大的“结局” 2.因为他不会输掉这场比赛（65 vs 59），所以他会选择13而不是14-这是策略-只是不会输掉（可以打平并赢得比赛）他可以计划自己的所有举动，因为他可以从乞讨中看到阵列，并且他不想以更少的举动来放松这场比赛

-amir从第一回合就知道下一步是什么，因为他不会输掉这场比赛（他可以以赢家的身份完成比赛，也可以以相同的tamara得分完成比赛）

-Amir可以预先计算游戏的完整动作树，以及Tamar如何响应每个动作，然后他如何响应每个动作。这种解决方案的问题可能是一棵大树（阿米尔和塔玛拉可以玩的不同游戏数量为2 ^ K-如果K更大，那么功能强大的计算机将花费数万亿年）。因此，在该游戏中需要诸如此类的有效解决方案。并要求阿米尔采取一些行动为自己制定策略

数组：

int[] array1 = {16,23,30,14,13,21,19,15};
TEST.coingame(array1);
System.out.println();

我的代码：

public static void coingame(int[] arr)
{ 
   int n = arr.length;  
   int i = 0, j = n-1,p1=0,p2=0,totalP2=0,totalP1=0; 

   while(j > i){

      if(arr[j]+arr[j-1]>arr[i]+arr[i+1]){
      p1 = arr[j];--j;
         if(arr[j]>arr[i]){
            p2=arr[j];--j;
         }else{  
            p2=arr[i];++i;
         } 
      }else{ 
         p1 = arr[i];++i;
         if(arr[j]>arr[i]){
            p2=arr[j];--j;
         }else{  
            p2=arr[i];++i;
         } 
      }

      System.out.println ("amir took "+p1);totalP1+=p1;
      System.out.println ("tamara took "+p2);totalP2+=p2;
   }

   System.out.println ("Final Score:");
   System.out.println ("amir total "+totalP1);
   System.out.println ("tamara total "+totalP2);
}

编辑：（Akshay Batra的答案）

public static int[] pickByAmir(int[] coins, int amirTook, int tamaraTook, int start, int end) {
    if(start>end) {
        int[] res = new int[2];
        res[0] = amirTook;
        res[1] = tamaraTook;
        return res;
    }
    int[] a = new int[2];
    a[0] = amirTook;
    a[1] = tamaraTook;
    if(coins.length==0)
        return a;
    amirTook = coins[start];
    coins = pickByTamara(coins, ++start , end);
    tamaraTook = coins[start];
    a = pickByAmir(coins, amirTook+a[0], tamaraTook+a[1], ++start, end);
    int[] b = new int[2];
    b[0] = amirTook;
    b[1] = tamaraTook;
    if(a[0]<a[1]){
        amirTook = coins[end];
        coins = pickByTamara(coins, start, --end);
        b = pickByAmir(coins, amirTook+b[0], tamaraTook+b[1], ++start, end);

        if(a[0]<b[0])
            return b;
    }
        System.out.println ("Amir took "+amirTook);
        System.out.println ("Tamara took "+tamaraTook);
    return a;
}
public static int[] pickByTamara(int[] coins, int start, int end){
    return coins[start] > coins[end] ? coins : swapArray(coins, start, end);
}

public static int[] swapArray(int[] coins, int start, int end) {
    int temp = coins[start];
    coins[start] = coins[end];
    coins[end] = temp;
    return coins;
}

public static void coingame(int[] arr) {
    int[] a = pickByAmir(arr, 0, 0, 0, arr.length-1);
    System.out.println ("Final Score: ");
    System.out.println ("Amir total: "+a[0]);
    System.out.println ("Tamara total: "+a[1]);
    }

Answer 1

此代码可与您的输入配合使用，并在不同的输入下进行尝试，看看是否中断（如果这样做），可以优化您的解决方案

通过调用方方法int[] a = pickByAmir(array1, 0, 0, 0, array1.length-1);

调用

a[0]将有amir的总数，而a[1]将具有tamara的总数。

int[] pickByAmir(int[] coins, int amirTook, int tamaraTook, int start, int end) {
    if(start>end) {
        int[] res = new int[2];
        res[0] = amirTook;
        res[1] = tamaraTook;
        return res;
    }
    int[] a = new int[2];
    a[0] = amirTook;
    a[1] = tamaraTook;
    if(coins.length==0)
        return a;
    amirTook = coins[start];
    coins = pickByTamara(coins, ++start , end);
    tamaraTook = coins[start];
    a = pickByAmir(coins, amirTook+a[0], tamaraTook+a[1], ++start, end);
    int[] b = new int[2];
    b[0] = amirTook;
    b[1] = tamaraTook;
    if(a[0]<a[1]){
        amirTook = coins[end];
        coins = pickByTamara(coins, start, --end);
        b = pickByAmir(coins, amirTook+b[0], tamaraTook+b[1], ++start, end);
        if(a[0]<b[0])
            return b;
    }
    return a;
}
int[] pickByTamara(int[] coins, int start, int end){
    return coins[start] > coins[end] ? coins : swapArray(coins, start, end);
}

int[] swapArray(int[] coins, int start, int end) {
    int temp = coins[start];
    coins[start] = coins[end];
    coins[end] = temp;
    return coins;
}

好运

Java中的两个角硬币游戏

1 个答案: