如何使用C#打印1到100而不进行任何循环

Question

我试图使用C#从1到100打印数字而不使用循环.有线索吗？

Answer 1

没有循环,没有条件,也没有硬编码的文字输出,又称"分而治之FTW"解决方案:

class P
{
    static int n;

    static void P1() { System.Console.WriteLine(++n); }

    static void P2() { P1(); P1(); }

    static void P4() { P2(); P2(); }

    static void P8() { P4(); P4(); }

    static void P16() { P8(); P8(); }

    static void P32() { P16(); P16(); }

    static void P64() { P32(); P32(); }

    static void Main() { P64(); P32(); P4(); }
}

替代方法:

using System;

class C
{
    static int n;

    static void P() { Console.WriteLine(++n); }

    static void X2(Action a) { a(); a(); }

    static void X5(Action a) { X2(a); X2(a); a(); }

    static void Main() { X2(() => X5(() => X2(() => X5(P)))); }
}

Answer 2

Console.Out.WriteLine('1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100');

Answer 3

可能是递归？

public static void PrintNext(i) {
    if (i <= 100) {
        Console.Write(i + " ");
        PrintNext(i + 1);
    }
}

public static void Main() {
    PrintNext(1);
}

Answer 4

多一个:

Console.WriteLine(
   String.Join(
      ", ", 
      Array.ConvertAll<int, string>(
         Enumerable.Range(1, 100).ToArray(), 
         i => i.ToString()
      )
   )
);

Answer 5

using static IronRuby.Ruby;

class Print1To100WithoutLoopsDemo
{
    static void Main() => 
      CreateEngine().Execute("(1..100).each {|i| System::Console.write_line i }");
}

嘿,为什么不呢？

Answer 6

Console.WriteLine('1');
Console.WriteLine('2');
...
Console.WriteLine('100');

...或者你会接受递归解决方案吗？

编辑:或者您可以这样做并使用变量:

int x = 1;
Console.WriteLine(x);
x+=1;
Console.WriteLine('2');
x+=1;
...
x+=1
Console.WriteLine('100');

Answer 7

Enumerable.Range(1, 100)
    .Select(i => i.ToString())
    .ToList()
    .ForEach(s => Console.WriteLine(s));

不确定这是否算作循环是隐藏的,但如果它是合法的,它是解决问题的惯用方法.否则你可以这样做.

    int count = 1;
top:
    if (count > 100) { goto bottom; }
    Console.WriteLine(count++);
    goto top;
bottom:

当然,这实际上是一个循环将被转换为什么,但是这些天写这样的代码肯定不赞成.

Answer 8

Enumerable.Range(1, 100).ToList().ForEach(i => Console.WriteLine(i));

以下是上述代码中发生的情况的细分:

• Enumerable.Range将指定的整数数字作为IEnumerator <Int32>返回
• Enumerable.ToList <T>将IEnumerable <T>转换为List <T>
• List <T> .ForEach采用lamdba函数并为列表中的每个项调用它

绩效考虑

ToList调用将导致为所有项目分配内存(在上面的示例中为100个int).这意味着O(N)空间复杂性.如果这是您的应用程序中的一个问题,即如果整数范围可能非常高,那么您应该避免直接列出和枚举项目.

不幸的是,ForEach不是开箱即用的IEnumerable扩展的一部分(因此需要在上面的例子中转换为List).幸运的是,这很容易创建:

static class EnumerableExtensions
{
    public static void ForEach<T>(this IEnumerable<T> items, Action<T> func)
    {
        foreach (T item in items)
        {
            func(item);
        }
    }
}

使用上面的IEnumerable扩展,现在在您需要将操作应用于IEnumerable的所有地方,您只需使用lambda调用ForEach即可.所以现在原始示例如下所示:

Enumerable.Range(1, 100).ForEach(i => Console.WriteLine(i));

唯一的区别是我们不再调用ToList,这会导致常量(O(1))空间使用...如果您处理的是大量项目,这将是一个非常明显的收益.

Answer 9

当我回答这个问题时,有人已经拥有了它,所以无论如何它都归功于Caleb:

void Main()
{
    print(0, 100);
}

public void print(int x, int limit)
{
    Console.WriteLine(++x);
    if(x != limit)
        print(x, limit);
}

Answer 10

没有循环,没有递归,只是一个类似哈希表的函数数组来选择如何分支:

using System;
using System.Collections.Generic;

namespace Juliet
{
    class PrintStateMachine
    {
        int state;
        int max;
        Action<Action>[] actions;

        public PrintStateMachine(int max)
        {
            this.state = 0;
            this.max = max;
            this.actions = new Action<Action>[] { IncrPrint, Stop };
        }

        void IncrPrint(Action next)
        {
            Console.WriteLine(++state);
            next();
        }

        void Stop(Action next) { }

        public void Start()
        {
            Action<Action> action = actions[Math.Sign(state - max) + 1];
            action(Start);
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            PrintStateMachine printer = new PrintStateMachine(100);
            printer.Start();
            Console.ReadLine();
        }
    }
}

Answer 11

只是为了丑陋的字面解释:

Console.WriteLine("numbers from 1 to 100 without using loops, ");

(你现在或以后可以笑,还是不笑)

Answer 12

我可以想到两种方式.其中一个涉及大约100行代码!

还有另一种方法可以多次重用一些代码而不使用while/for循环...

提示:创建一个打印从1到N的数字的函数.应该很容易使其适用于N = 1.然后考虑如何使其适用于N = 2.

Answer 13

使用正则表达式

using System.Text.RegularExpressions;

public class Hello1
{
   public static void Main()
   {

      // Count to 128 in unary
      string numbers = "x\n";
      numbers += Regex.Replace(numbers, "x+\n", "x$&");
      numbers += Regex.Replace(numbers, "x+\n", "xx$&");
      numbers += Regex.Replace(numbers, "x+\n", "xxxx$&");
      numbers += Regex.Replace(numbers, "x+\n", "xxxxxxxx$&");
      numbers += Regex.Replace(numbers, "x+\n", "xxxxxxxxxxxxxxxx$&");
      numbers += Regex.Replace(numbers, "x+\n", "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx$&");
      numbers += Regex.Replace(numbers, "x+\n", "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx$&");

      // Out of 1..128, select 1..100
      numbers = Regex.Match(numbers, "(.*\n){100}").Value;

      // Convert from unary to decimal
      numbers = Regex.Replace(numbers, "x{10}", "<10>");
      numbers = Regex.Replace(numbers, "x{9}", "<9>");
      numbers = Regex.Replace(numbers, "x{8}", "<8>");
      numbers = Regex.Replace(numbers, "x{7}", "<7>");
      numbers = Regex.Replace(numbers, "x{6}", "<6>");
      numbers = Regex.Replace(numbers, "x{5}", "<5>");
      numbers = Regex.Replace(numbers, "x{4}", "<4>");
      numbers = Regex.Replace(numbers, "x{3}", "<3>");
      numbers = Regex.Replace(numbers, "x{2}", "<2>");
      numbers = Regex.Replace(numbers, "x{1}", "<1>");
      numbers = Regex.Replace(numbers, "(<10>){10}", "<100>");
      numbers = Regex.Replace(numbers, "(<10>){9}", "<90>");
      numbers = Regex.Replace(numbers, "(<10>){8}", "<80>");
      numbers = Regex.Replace(numbers, "(<10>){7}", "<70>");
      numbers = Regex.Replace(numbers, "(<10>){6}", "<60>");
      numbers = Regex.Replace(numbers, "(<10>){5}", "<50>");
      numbers = Regex.Replace(numbers, "(<10>){4}", "<40>");
      numbers = Regex.Replace(numbers, "(<10>){3}", "<30>");
      numbers = Regex.Replace(numbers, "(<10>){2}", "<20>");
      numbers = Regex.Replace(numbers, "(<[0-9]{3}>)$", "$1<00>");
      numbers = Regex.Replace(numbers, "(<[0-9]{2}>)$", "$1<0>");
      numbers = Regex.Replace(numbers, "<([0-9]0)>\n", "$1\n");
      numbers = Regex.Replace(numbers, "<([0-9])0*>", "$1");

      System.Console.WriteLine(numbers);

   }
}

输出:

# => 1
# => 2
# ...
# => 99
# => 100

Answer 14

方法A:

Console.WriteLine('1');
Console.WriteLine('print 2');
Console.WriteLine('print 3');
...
Console.WriteLine('print 100');

方法B:

func x (int j)
{
  Console.WriteLine(j);
  if (j < 100)
     x (j+1);
}

x(1);

Answer 15

只是LINQ吧......

Console.WriteLine(Enumerable.Range(1, 100)
                            .Select(s => s.ToString())
                            .Aggregate((x, y) => x + "," + y));