均匀地将集合分配到新的阵列中

Question

我正在尝试创建一种方法,将数组均匀分布到X个新数组中,其中只允许15个项目pr数组,并且只允许创建一个新数组,前者有10个项目,除非该数组少于10项.

编辑

让我的问题对未来的读者更容易理解.

• 这就像一个面料.
• 您需要构建X个产品.
• 一个产品需要为员工构建T金额.

您需要多少员工,以及如何分担他们之间的工作量？

结束编辑

 Max allowed number in array = 15;
 Min allowed number in array = 10;    

 Number = Numbers of Items in the Collection.

 Number  5 =>   [5]
 Number 13 =>   [13]
 Number 16 =>   [10] [6]
 Number 29 =>   [15] [14]
 Number 30 =>   [15] [15]
 Number 31 =>   [11] [10] [10]
 Number 32 =>   [12] [10] [10]
 Number 33 =>   [11] [11] [11]

我试图在C#中解决这个问题.

到目前为止,这是我的代码,但它在16 = [16],29 = [19] [10],38 = [18] [10] [10]等数字时失败了

const int maxAllowedOrderLines = 15;
const int minAllowedOrderLines = 10;
var optimalOrderDisp = new List<int>();

Console.WriteLine("Number of OrderLines");
int linjer = Convert.ToInt32(Console.ReadLine());

if (linjer <= maxAllowedOrderLines)
   optimalOrderDisp.Add(linjer);
else
{
   for (var i = maxAllowedOrderLines; i > 0; i--)
   {
      var maxOrderLines = linjer%i;
      if (maxOrderLines == 0 || i <= minAllowedOrderLines || linjer < maxAllowedOrderLines)
      {
         Console.WriteLine("Optimal number of order lines {0}--{1}", i, (double) linjer/(double) i);

          var optimalNumberOfOrders = linjer/i;
          for (var orderNumber = 0; orderNumber < optimalNumberOfOrders; orderNumber++)
          {
             optimalOrderDisp.Add(i);
          }

          if (maxOrderLines != 0)
             optimalOrderDisp[0] += maxOrderLines;
          break;
       }
    }
 }
 foreach (var i1 in optimalOrderDisp)
 {
    Console.Write("[{0}]", i1);
 }
 Console.WriteLine();

Answer 1

呃......

const double bucketSize = 15.0;
var totalItems = (double)linjer;
var optimumBuckets = Math.Ceiling(totalItems / bucketSize);
var itemsPerBucket = (int)Math.Ceiling(totalItems / optimumBuckets);

var buckets = new int[(int)optimumBuckets];

var itemsLeft = (int)totalItems
for (var i = 0; i < buckets.length; i++)
{
    if (itemsLeft < itemsPerBucket)
    {
        buckets[i] = itemsLeft;
    }
    else
    {
        buckets[i] = itemsPerBucket;
    }
    itemsLeft -= itemsPerBucket;
}

似乎做你想要的.