Question

说我想将y个项目均匀地分发到x个桶。如果x是y的倍数，则此分布将是均匀的，如果不是，我最终可以在每个存储桶中使用0个项目。例如：

例如：我有3个桶，我想分发2个项目。由于进行除法(2/3)将导致每个桶0个项目。我如何实现1，1，0的分发？

Answer 1

第一个y mod x存储分区将包含(y div x) + 1个项目，其余部分将包含y div x个项目

Answer 2

这种想法应该有效：

package sandbox;

public class Sandbox
{

    public static void main(String[] args)
    {
        int numBuckets = 12;
        int numItems = 34;

        int itemsPerBucket = (numItems / numBuckets);
        int remainingItems = (numItems % numBuckets);

        for (int i = 1; i <= numBuckets; i++)
        {
            int extra = (i <= remainingItems) ? 1:0;
            System.out.println("bucket " + i + " contains " + (itemsPerBucket + extra) + " items.");
        }
    }
}

输出：

bucket 1 contains 3 items.
bucket 2 contains 3 items.
bucket 3 contains 3 items.
bucket 4 contains 3 items.
bucket 5 contains 3 items.
bucket 6 contains 3 items.
bucket 7 contains 3 items.
bucket 8 contains 3 items.
bucket 9 contains 3 items.
bucket 10 contains 3 items.
bucket 11 contains 2 items.
bucket 12 contains 2 items.

请注意，您所做的唯一循环是讨论每个存储桶。你可以轻松地询问一个桶号，看看它中有多少物品没有循环！

Answer 3

每个桶获得y / x（整数除法）项，y％x桶获得1个额外项。

Answer 4

你的问题是模糊的，但是从itemsNumber的分区中取出余数（如果它们在列表或数组中可能是索引），那么作为桶的索引的桶数将给出你正在寻找的均匀分布。

int bucketIndex = itemNumber % numberOfBuckets;

Answer 5

它看起来你知道如何解决编程问题，但你正在寻找一个想法，我认为你在pascal的三角形中的答案，认为与pascal的三角形有点不同，把你的整个项目放到三角形的顶部单元格中。所以，实际上，我认为这不是最好的努力

Answer 6

这里是一个递归解决方案，只是为了好玩。

import java.util.Arrays;

public class PizzaDistributor {

  public int[] distributePizzaToEatersEvenly(int numSlices, int[] eaters) {
    if (eaters == null || eaters.length == 0) {
      return null;
    }
    if (eaters.length > numSlices) {
      for (int i = 0; i < numSlices; i++) {
        eaters[i] += 1;
      }
    } else {
      int slices = numSlices / eaters.length;
      for (int i = 0; i < eaters.length; i++) {
        eaters[i] = slices;
      }
      //If there is a remainder to distribute, distribute it evenly, recursively.
      if ((numSlices % eaters.length) > 0) {
        distributePizzaToEatersEvenly(numSlices % eaters.length, eaters);
      }
    }
    return eaters;
  }

  public static void main(String[] args){
    PizzaDistributor pizzaDistributor = new PizzaDistributor ();
    System.out.println("0 slices, 1 eater:" +Arrays.toString(pizzaDistributor.distributePizzaToEatersEvenly(0,new int[1])));
    System.out.println("0 slices, 2 eaters:" +Arrays.toString(pizzaDistributor.distributePizzaToEatersEvenly(0,new int[2])));
    System.out.println("0 slices, 0 eaters:" +Arrays.toString(pizzaDistributor.distributePizzaToEatersEvenly(0,new int[0])));
    System.out.println("10 slices, 0 eaters:"+Arrays.toString(pizzaDistributor.distributePizzaToEatersEvenly(10,new int[0])));
    System.out.println("10 slices, 1 eater:" +Arrays.toString(pizzaDistributor.distributePizzaToEatersEvenly(10,new int[1]))); //lucky guy!
    System.out.println("10 slices, 3 eaters:" +Arrays.toString(pizzaDistributor.distributePizzaToEatersEvenly(10,new int[3])));
    System.out.println("10 slices, 5 eaters:" +Arrays.toString(pizzaDistributor.distributePizzaToEatersEvenly(10,new int[5])));
    System.out.println("10 slices, 7 eaters:" +Arrays.toString(pizzaDistributor.distributePizzaToEatersEvenly(10,new int[7])));
    System.out.println("10 slices, 10 eaters:" +Arrays.toString(pizzaDistributor.distributePizzaToEatersEvenly(10,new int[10])));
    System.out.println("10 slices, 13 eaters:" +Arrays.toString(pizzaDistributor.distributePizzaToEatersEvenly(10,new int[13])));
    System.out.println("10 slices, 20 eaters:" +Arrays.toString(pizzaDistributor.distributePizzaToEatersEvenly(10,new int[20])));

  }

}

产生此输出：

0 slices, 1 eater:[0]
0 slices, 2 eaters:[0, 0]
0 slices, 0 eaters:null
10 slices, 0 eaters:null
10 slices, 1 eater:[10]
10 slices, 3 eaters:[4, 3, 3]
10 slices, 5 eaters:[2, 2, 2, 2, 2]
10 slices, 7 eaters:[2, 2, 2, 1, 1, 1, 1]
10 slices, 10 eaters:[1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
10 slices, 13 eaters:[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0]
10 slices, 20 eaters:[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

Process finished with exit code 0

平均分配桶中的“项目”（尽力而为）

6 个答案: