Question

给出以下输入

10 4 3 5 5 7

Where

10 = Total Score

4 = 4 players

3 = Score by player 1
5 = Score by player 2
5 = Score by player 3
7 = Score by player 4

输出应该打印得分等于10的玩家的索引。因此，对于给定的上述输出，它应该打印1 4或2 3，因为玩家1 +玩家4得分加起来10，所以得分也是播放器2和播放器3.我不需要打印两种或所有组合。我只想打印任何一种有效的组合。

For INPUT : 8 3 2 2 4 OUPUT : 1 2 3 since scores of player 1 player 2 and player 3 equal the total score of 8

所以我一直在阅读过去一周的动态编程教程和视频，并获得有关堆栈溢出的帮助以修复我的初始代码。

以下是我到目前为止

    public boolean findSolution(int[] scores, int total) {
    int W = total;
    int players = scores.length;

    boolean[][] myArray = new boolean[players + 1][total + 1];

    for (int player = 0; player <= players; player++) {
        myArray[player][0] = true;
    }
    for (int score = 1; score < total; score++) {
        myArray[0][score] = false;
    }
    for (int player = 1; player <= players; player++) {
        for (int score = 1; score <= total; score++) {
            myArray[player][score] = myArray[player - 1][score];
            if (score >= scores[player - 1]) {
                myArray[player][score] = myArray[player - 1][score
                        - scores[player - 1]]
                        || myArray[player][score];
            }
        }
    }
    return myArray[players][W];

}

此逻辑创建2d数组并进行穷举搜索以查看给定总数的组合是否可行，如果是，则返回true，否则返回false。现在我无法打印出真实的玩家的索引。如果有人可以帮我打印一组得分等于总数的球员的索引，那将非常感激。我不需要打印所有组合。

如果您不理解，我也不会问任何问题，因为我不是以英语为母语的人。

Answer 1

好的，所以在你创建并更新了布尔数组myArray之后。

我们将从最后一个玩家迭代到第一个玩家，检查我们是否可以在最终结果中使用当前玩家

int currentScore = total;//Maintain the current score.
for(int i = lastPlayer ; i >= 0; i--){

}

在for循环中，为了检查当前i玩家是否属于我们的最终玩家，我们需要检查是否存在currentScore - score of i player的解决方案

if (currentScore >= scores[i] && (i == 0 || myArray[i - 1][currentScore - scores[i]]) {
      //Update current score      
      currentScore -= scores[i];
      //Print name of the player.
      System.out.println("Player " + i);              
}

动态编程以解决给定和的子集

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