如何计算组合给予时的指数（词典编纂顺序） [英] How to calculate the index (lexicographical order) when the combination is given

问题描述

我知道，有一个算法，允许，给定数（不重复，没有秩序）的组合，计算字典顺序的索引。
这将是非常有益的我的应用程序加速比的东西...

I know that there is an algorithm that permits, given a combination of number (no repetitions, no order), calculates the index of the lexicographic order.
It would be very useful for my application to speedup things...

例如：

combination(10, 5)  
1 - 1 2 3 4 5  
2 - 1 2 3 4 6  
3 - 1 2 3 4 7  
....  
251 - 5 7 8 9 10  
252 - 6 7 8 9 10  

我需要，该算法返回给定的组合的索引。
ES：指数（2，5，7，8，10） - >首页

I need that the algorithm returns the index of the given combination.
es: index( 2, 5, 7, 8, 10 ) --> index

编辑：其实我使用的Java应用程序生成的所有组合C（53，5），并将其插入到一个TreeMap的。 我的想法是创建一个包含所有组合（及相关数据），我可以使用索引算法的数组。
一切都是用来加快联合搜索。 不过，我尝试了一些（不是全部）的解决方案，以及您提出的算法是慢了一个get（）从TreeMap中。
如果有帮助：我需要的是5从53开始，从0到52的组合

actually I'm using a java application that generates all combinations C(53, 5) and inserts them into a TreeMap. My idea is to create an array that contains all combinations (and related data) that I can index with this algorithm.
Everything is to speedup combination searching. However I tried some (not all) of your solutions and the algorithms that you proposed are slower that a get() from TreeMap.
If it helps: my needs are for a combination of 5 from 53 starting from 0 to 52.

再次感谢所有： - ）

Thank you again to all :-)

推荐答案

下面是一个片段，就做好了。

Here is a snippet that will do the work.

#include <iostream>

int main()
{
    const int n = 10;
    const int k = 5;

    int combination[k] = {2, 5, 7, 8, 10};

    int index = 0;
    int j = 0;
    for (int i = 0; i != k; ++i)
    {
        for (++j; j != combination[i]; ++j)
        {
            index += c(n - j, k - i - 1);
        }
    }

    std::cout << index + 1 << std::endl;

    return 0;
}

它假设你有一个函数

It assumes you have a function

int c(int n, int k);

这将返回的选择k个元素出n个元素的组合的数目。 循环计算的组合preceding的给定组合的数目。 通过添加一个在最后我们得到的实际指数。

that will return the number of combinations of choosing k elements out of n elements. The loop calculates the number of combinations preceding the given combination. By adding one at the end we get the actual index.

对于给定的组合有 C（9,4）=含1和126组合，因此preceding它在字典顺序。

For the given combination there are c(9, 4) = 126 combinations containing 1 and hence preceding it in lexicographic order.

含2作为最小号的组合中有

Of the combinations containing 2 as the smallest number there are

C（7,3）= 35的组合具有3作为第二最小的数

c(7, 3) = 35 combinations having 3 as the second smallest number

C（6,3）= 20的组合具有4作为第二最小的数

c(6, 3) = 20 combinations having 4 as the second smallest number

所有这些都是preceding给定的组合

All of these are preceding the given combination.

含2和5作为两个最小数字的组合中有

Of the combinations containing 2 and 5 as the two smallest numbers there are

C（4,2）=具有6作为第三最小数字6的组合

c(4, 2) = 6 combinations having 6 as the third smallest number.

所有这些都是preceding给定的组合

All of these are preceding the given combination.

等

如果你把一个print语句在内部循环，你会得到的数字 126，35，20，6，1。 希望这解释了code。

If you put a print statement in the inner loop you will get the numbers 126, 35, 20, 6, 1. Hope that explains the code.

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