排列 [英] permutations

问题描述

我正在做一个小项目，对编程还是陌生的.

我需要做一个清单，在该清单中，我可以烫发12个中的4个部分中的12个中的任意1个，而且我不知道从哪里开始.

谁能帮忙.

亲切的问候
jim

Hi,

I am doing a small project and I am new to programming.

I need to do a list where I can perm any 1 from 12 from 4 sections of 12 and I do not have a clue where to start.

Can anyone assist please.

kind regards
jim

推荐答案

您应该使用STL容器和算法，但需要学习它...
您的马匹应该放在 std :: list< int> [^ ]，然后使用一种算法功能，可能是生成 [

You should use STL containers and algorithms, but you need to learn it...
Your horses should be in an std::list< int >[^], then use one of algorithm function, may be generate[^] with your lambda function as third parameter.

听起来像是要组合而不是排列.无需使用STL容器.您可以尝试类似的方法，我用它来确认用牛顿-吉拉德(Newton-Girard)进行默认计算的第k次.

基本上，创建一个编号系统并使用严格增加的#遍历有序集.

It sounds like you want combinations and not permutations. There is no need to use an STL container. You can try something like this, which I utilized to confirm a k-th to default computation I made with Newton-Girard.

Basically, create a numbering system and iterate over the ordered sets with strictly increasing #''s.

namespace {
    void print(int *s, int const & k) {
        for (int j=0; j<k; j++)
            printf("%i ", s[j]+1);
        printf("\n");
    }
    int diff_sum(int *s, int const & k) {
        int sum=0;
        for (int j=1; j<k; j++)
            sum+=s[j]-s[j-1];
        return sum;
    }
};

void combination_list(int const & N, int const & k) {
    int selection[k], reset[k-1];
    selection[0]=N-k;
    int i;
    for (i=1; i<k; i++)
        selection[i]=reset[i-1]=N-k+i;
    int count=1;
    print(selection, k);
    int j;
    while (selection[0]+diff_sum(selection, k)>=k) {
        for (j=0; j<k; j++) {
            if ( diff_sum(&selection[j], k-j)==k-1-j) {
                selection[j]-=1;
                for (i=j+1; i<k; i++)
                    selection[i]=reset[i-1];
                count++;
                print(selection, k);
            }
        }
    }
    printf("count = %i\n", count);
}

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