基于位置计算组合 [英] Calculate Combination based on position

问题描述

创建给定的字符集合C的函数，可以生成第N个组合或返回该系列定开始位置（NS）的组合和结束位置（氖）和组合（MX）的最大长度。

Create a function that given a character set C, can generate the Nth combination OR return the series of combination given a starting position (Ns) and ending position (Ne) and the maximum length of the combination (Mx).

一个具体的例子：

让C = [A，B，C]

Let C = [A,B,C]

我们知道，不同的组合看起来像下面的假设的Mx = 3 （组合将是不同长度不同）的：

We know that different combinations would look like the following assuming Mx = 3 (the combination would be different for different lengths):

1. AAA
2. AAB
3. AAC
4. ABA
5. ABB
6. ABC
N. ... Etc

如果我们是通过下列参数：

If we was to pass the following parameters :

C = [A，B，C]的Mx = 3-NS = 3 NE = 3

C = [A,B,C] Mx = 3 Ns = 3 Ne = 3

我们希望以下结果：

AAC

如果我们是通过下列参数：

If we was to pass the following parameters :

C = [A，B，C]的Mx = 3-NS = 4 NE = 6

C = [A,B,C] Mx = 3 Ns = 4 Ne = 6

我们希望以下结果：

4. ABA
5. ABB
6. ABC

有关的方案中，编程语言是不相关的。但是C＃是preferred。另外最重要的是如何解决它的解释。

For the solution, the programming language is not relevant. However C# would be preferred. Also most important would be an explanation of how its solved.

我期待着惊人的宗师的堆栈溢出...

I look forward to the amazing Guru's of Stack Overflow...

推荐答案

由于指数的 N （从0开始）到的 N 符号组合的顺序，你可以在我个符号，通过计算的 N / N ^我 ％<强> N （使用整数除法和余）

Given an index N (0-based) into the sequence of combinations of n symbols, you can get the i'th symbol by calculating N / nⁱ % n (using integer division and remainder)

例如：

C = {A, B, C} (giving n = 3)
N = 6
i = 0 => 6 / 3^0 % 3 = 0 (symbol 0 = A)
i = 1 => 6 / 3^1 % 3 = 2 (symbol 2 = C)
i = 2 => 6 / 3^2 % 3 = 0 (symbol 0 = A)
Resulting sequence: ACA

的序列被视为一个碱基的 N 号，以及各个单独数字计算

The sequence is treated as a base-n number, and the individual digits are calculated.

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