在阅读了一些关于递归函数的多嵌套循环的主题后.我得到了最终的解决方案，我从其他线程复制了一些行，但添加了一些过滤数据的条件.下面是条形组合切割的代码和答案，以满足不再浪费1‘，但我的最终目标是https://en.wikipedia.org/wiki/Cutting_stock_problem.

    private static double[] BarCuts = { 6.2, 5, 3, 4};  // List of Bar Cuttings.
    static void Main(string[] args)
    {
        List<int[]> collection = new List<int[]>();
        int[] limits = GetLimits(BarCuts); 
        printHading(BarCuts);
        
        AddToCollectionRecursive(collection, limits);
        
        foreach (var item in collection)
            printResult(item); 
        Console.ReadLine();
    }

    static void AddToCollectionRecursive(List<int[]> collection, params int[] limits)
    {
        AddTo(collection, new List<int>(), limits, limits.Length - 1);
    }

    static void AddTo(List<int[]> collection, IEnumerable<int> value, IEnumerable<int> counts, int left)
    {
        for (var i = 0; i < counts.First(); i++)
        {
            var list = value.ToList();
            list.Add(i);
            if (left == 0)
            {
                double tl = Combination_Length(list);
                if (tl > 20) { break; }

                double waste = 20 - tl;          // 20 = avlb bar length.
                if (waste < 1)
                    collection.Add(list.ToArray());
            }
            else
            {
                AddTo(collection, list, counts.Skip(1), left - 1);
            }
        }
    }

    static double Combination_Length(List<int> CutsQty)
    {
        double tl = 0;
        for (int j = 0; j < CutsQty.Count; j++)
            tl += CutsQty[j] * BarCuts[j];
        return tl;
    }

答案如下.

A = 6.2 , B = 5 , C = 3 , D = 4

5D = TL= 20，废物= 0.00

4C&；2D=>；TL=20，废物=0.00

1B&；1C&；3D=>；TL=20，废物=0.00

1B&A；5C=>；TL=20，废物=0.00

2B&；2C&；1D=>；TL=20，浪费=0.00

4B=>；TL=20，废物=0.00

1A&；3C&；1D=>；TL=19.2，废物=0.80

1A&；1B&；2D=>；TL=19.2，废物=0.80

1A 2B 1C = TL= 19.2，废物= 0.80

2a&；1C&；1D=>；TL=19.4，废物=0.60