假设我想将y 项目平均分配到x 桶。如果x 是y 的倍数，则此分布将是均匀的，否则我可以在每个存储桶中得到0 项目。例如：
例如：我有 3 存储桶，我想每个分配 2 项目。由于进行除法(2/3) 将导致每个桶的0 项目。如何实现，1、1、0 的分布？
【问题讨论】：
标签： java algorithm math distribution
这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])));
}
}
产生这个输出：
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
【讨论】：
看起来你知道如何解决编程问题，但你正在寻找一个想法，我认为你在帕斯卡三角形中的答案，与帕斯卡三角形有点不同，将你的整个项目放入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
【讨论】：
每个桶得到 y/x（整数除法）项目，y % x 个桶得到 1 个额外项目。
【讨论】：
这种想法应该可行：
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.");
}
}
}
这个的输出：
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。你可以很容易地问一个桶号，看看里面有多少项目没有循环！
【讨论】：
第一个 y mod x 存储桶将有 (y div x) + 1 项目，其余的将有 y div x 项目
【讨论】：
您的问题有点含糊，但是从 itemNumber 的除法中取出余数（如果它们在列表或数组中，这可能是索引）和存储桶的数量作为存储桶的索引会给您您正在寻找的均匀分布。
int bucketIndex = itemNumber % numberOfBuckets;
【讨论】：