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有没有没有没有没有if/else/switch/的公式?可替代解决方案[,]矩阵的ETC

math c#

Caspar Kleijne · 技术社区 · 15 年前

有没有没有没有没有if/else/switch/的公式?可以替代解决方案[,]矩阵的ETC,这对性能/效率有什么影响或区别?

     class Program
{
    private static Random r = new Random();
    // names of the "moves"
    private static string[] rps = { "PAPER", "ROCK", "SCISSORS"};
    // result feedback string to be formatted
    private static string[] feedback = { "{1} Beats {0}, You Loose!","{0} Equals {1}, Draw!",  "{0} Beats {1}, You Win!" };
    // solution matrix ( 0 = loose ; 1 = draw ; 2 = win  // array1: for paper; array2: for rock; array3: for scissors; )
    private static int[,] solution = {{1, 2, 0},{0, 1, 2},{2, 0, 1}};

    /// <summary>
    /// Rock Paper scissors solution w/o calculation or if/case/else/
    /// </summary>
    /// <param name="args">dummy.</param>
    static void Main(string[] args)
    {
            // simulate the players move
            int player = r.Next(3);

            // simulate the computers move
            int computer = r.Next(3);

            // retrieve the result from the matrix
            int result = solution[player, computer];

            //write the result of the match
            Console.WriteLine(String.Format("you : {0} vs {1} : computer", rps[player], rps[computer]));
            Console.WriteLine(String.Format(feedback[result], rps[player], rps[computer]));

    }
}

1 回复 | 直到 15 年前

Guffa 15 年前

是的,有一个非常简单的公式:

player computer  solution  (c+4-p)%3
  0       0         1          1
  0       1         2          2
  0       2         0          0
  1       0         0          0
  1       1         1          1
  1       2         2          2
  2       0         2          2
  2       1         0          0
  2       2         1          1

所以你可以使用:

int result = (computer + 4 - player) % 3;

访问数组并计算值需要花费大量的时间。但是,此应用程序的性能差异可以忽略不计。将结果写入控制台比使用数组或计算值花费的时间要长得多。当然,通过计算不需要数组的值,但由于数组太小,所以没有什么区别。

还要考虑解决方案的可读性。这个公式与你使用它的目的没有逻辑上的联系,它只是一种获得特定结果的方法,所以你需要一个大的注释来解释它完成了什么……

编辑:

如果您希望关注可读性,可以将逻辑放在单独的类中:

public class Play {

  public enum Value { Paper = 0, Rock = 1, Scissors = 2 }

  private Value _value;

  public Play(Random rnd) {
    _value = (Value)rnd.Next(3);
  }

  public bool SameAs(Play other) {
    return _value == other._value;
  }

  public bool Beats(Play other) {
    return
      (_value == Value.Paper && other._value == Value.Rock) ||
      (_value == Value.Rock && other._value == Value.Scissors) ||
      (_value == Value.Scissors && other._value == Value.Paper);
  }

  public override string ToString() {
    switch (_value) {
      case Value.Paper: return "PAPER";
      case Value.Rock: return "ROCK";
      default: return "SCISSORS";
    }
  }

}

现在逻辑变得更清晰了:

Random r = new Random();

Play player = new Play(r);
Play computer = new Play(r);

Console.WriteLine("you : {0} vs {1} : computer", player, computer);

string feedback;
if (player.SameAs(computer)) {
  feedback = "{0} Equals {1}, Draw!";
} else if (player.Beats(computer)) {
  feedback = "{0} Beats {1}, You Win!";
} else {
  feedback = "{1} Beats {0}, You Loose!";
}

Console.WriteLine(feedback, player, computer);