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Pavel
Asked:2024-08-26 17:47:31 +0000 UTC2024-08-26 17:47:31 +0000 UTC

C#：一行解决方案和 While 循环

  • 772

在控制台输入时，用户输入从 0 到 long 左右的任何数字。Max 该数字放置在秤的左盘上。三者的力量需要平衡这些天平。嗯，就是如果输入的是7，那么我们在7的左边放3，右边放9和1，权重不能重复。也就是说，您不能两次输入 1。

目标是用最短的线路解决这个问题。它有多难读并不重要。

到目前为止，我已经能够得到这个三行解决方案：

List<List<long>> weights = new List<List<long>> { new List<long> { long.Parse(Console.ReadLine()) }, new List<long>() };
while (weights[0].Sum() != weights[1].Sum()) weights[weights[0].Sum() > weights[1].Sum() ? 1 : 0].Add((long)Math.Pow(3, ((Func<long, long, bool>)((number, power) => number > (long)((Math.Pow(3, power) - 1) / 2)))(weights[0].Sum() > weights[1].Sum() ? weights[0].Sum() - weights[1].Sum() : weights[1].Sum() - weights[0].Sum(), ((Func<long, long>)(number => (long)Math.Ceiling(Math.Log(number, 3))))(weights[0].Sum() > weights[1].Sum() ? weights[0].Sum() - weights[1].Sum() : weights[1].Sum() - weights[0].Sum())) ? ((Func<long, long>)(number => (long)Math.Ceiling(Math.Log(number, 3))))(weights[0].Sum() > weights[1].Sum() ? weights[0].Sum() - weights[1].Sum() : weights[1].Sum() - weights[0].Sum()) : ((Func<long, long>)(number => (long)Math.Ceiling(Math.Log(number, 3))))(weights[0].Sum() > weights[1].Sum() ? weights[0].Sum() - weights[1].Sum() : weights[1].Sum() - weights[0].Sum()) - 1));
Console.WriteLine(JsonSerializer.Serialize(weights));

我相信你想要的可以使用 Linq 的 Agreggate 方法或使用 lambda 来实现，但是 while 的存在非常令人困惑。是否可以用它做点什么？

这是我最初的解决方案，我试图将其压缩为一行：

 class Program
 {
     static void Main(string[] args)
     {
             long weight = long.Parse(Console.ReadLine());
             List<List<long>> weights = new List<List<long>>
             {
                 new List<long>(),
                 new List<long>()
             };
             weights[0].Add(weight);
             FindThreeDegreeWeights(weights);
             Console.WriteLine(JsonSerializer.Serialize(weights));            
     }

     static long MinPowerOfThreeNotLessThan(long number)
     {
         long power = (long)Math.Ceiling(Math.Log(number, 3));
         return power;
     }

     static bool IsNumberGreaterThanSumOfPreviousPowersOfThree(long number, long power)
     {
         long sum = (long)((Math.Pow(3, power) - 1) / 2);
         return number > sum;
     }

     static void FindThreeDegreeWeights(List<List<long>> weights)
     {
         long leftSum = weights[0].Sum();
         long rightSum = weights[1].Sum();

         if (leftSum == rightSum)
             return;

         if (leftSum > rightSum)
         {
             long difference = leftSum - rightSum;
             long upperDegree = MinPowerOfThreeNotLessThan(difference);
             if (IsNumberGreaterThanSumOfPreviousPowersOfThree(difference, upperDegree))
             {
                 weights[1].Add((long)Math.Pow(3, upperDegree));
             }
             else
             {
                 weights[1].Add((long)Math.Pow(3, upperDegree - 1));
             }
         }
         else
         {
             long difference = rightSum - leftSum;
             long upperDegree = MinPowerOfThreeNotLessThan(difference);
             if (IsNumberGreaterThanSumOfPreviousPowersOfThree(difference, upperDegree))
             {
                 weights[0].Add((long)Math.Pow(3, upperDegree));
             }
             else
             {
                 weights[0].Add((long)Math.Pow(3, upperDegree - 1));
             }
         }
         FindThreeDegreeWeights(weights);
     }
 }
c#

2 个回答

  1. 也许它会有用：在Python中转换为平衡三元系统的整数算法（删除了零）

    def TriEqSystem(n):
    lst = []
    p = 1
    while n:
        n, m = divmod(n, 3)
        if m:
            lst.append(p*(3-2*m))  #добавляет p или (-p) для остатков 1 и 2
            n += (m-1)             #увеличить следующий разряд для остатка 2
        p *= 3
    return lst

for i in range(1,101):
    l = TriEqSystem(i)
    print(i, sum(l), l)

...
10 10 [1, 9]
11 11 [-1, 3, 9]
12 12 [3, 9]
13 13 [1, 3, 9]
14 14 [-1, -3, -9, 27]
15 15 [-3, -9, 27]
16 16 [1, -3, -9, 27]
17 17 [-1, -9, 27]
...
    • 2
  2. Best Answer

    更新

    这是使用 @MBo 算法的单行解决方案（但我自己想出了）：

    using System;
static class Program {
    static void Main() {
        for (long i = 1, n = Int64.Parse(Console.ReadLine()), j = n % 3; n > 0; i += i << 1, n = n / 3 + (j == 0 ? 0 : j - 1), j = n % 3 ) if (j > 0) Console.WriteLine(i * ((j << 1) - 3));
    }
}

    如果答案中的三的幂为正，则将其与数字一起放置。

    也可以使用 Linq：

    using System;
using System.Linq;
static class Program {
    static void Main() {
        foreach (var m in Enumerable.Range(Int64.Parse(Console.ReadLine()!) is long N && 1L is long i? 0 : 0, 40)
            .Select(s => new { x = N = Math.DivRem(N, 3, out long j), Rem = j, y = j == 0 ? N : N += j - 1 })
            .Where(w => w.Rem > 0).Select(s => (i *= 3) / 3 * ((s.Rem << 1) - 3))) Console.WriteLine(m);
    }
}
    • 2

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