将给定数量M随机分成N个部分

Question

因此,我的想法是能够将2.00美元分成10个人,并且每个人将随机获得$ x.xx金额.(N和M将始终限制为2位小数且> 0)

例:{0.12,0.24,1.03,0.01,0.2,0.04,0.11,0.18,0.05,0.02}

目前我尝试过:

private static BigDecimal[] randSum(int n, double m)
{
    Random rand = new Random();
    BigDecimal randNums[] = new BigDecimal[n], sum = new BigDecimal(0).setScale(2);

    for (int i = 0; i < randNums.length; i++)
    {
        randNums[i] = new BigDecimal(rand.nextDouble()).setScale(2, RoundingMode.HALF_EVEN);
        sum = sum.add(randNums[i]);
    }

    for (int i = 0; i < randNums.length; i++)
    {
        BigDecimal temp1 = randNums[i].divide(sum, 2, RoundingMode.HALF_EVEN);
        BigDecimal temp2 = temp1.multiply(new BigDecimal(m).setScale(2));
        randNums[i] = temp2;
    }

    return randNums;
}

public static void main(String[] args)
{
    BigDecimal d[] = randSum(5, 2);

    double sum = 0;
    for (BigDecimal n : d)
    {
        sum += n.doubleValue();
        System.out.println(n);
    }
    System.out.println("total: " + sum);
}

但BigDecimals太混乱了,他们没有加起来.有时总数是1.98或2.01.由于Double-precision浮点数,双打不起作用.

代码取自:

获得总和为M的N个随机数

Answer 1

假设你需要一个固定的精度(作为prec参数传递):

static public BigDecimal[] split(BigDecimal sum, int prec, int count) {
    int s = sum.scaleByPowerOfTen(prec).intValue();
    Random r = new Random();
    BigDecimal[] result = new BigDecimal[count];
    int[] v = new int[count];

    for (int i = 0; i < count - 1; i++)
       v[i] = r.nextInt(s);
    v[count - 1] = s;

    Arrays.sort(v);
    result[0] = BigDecimal.valueOf(v[0]).scaleByPowerOfTen(-prec);
    for (int i = 1; i < count; i++)
       result[i] = BigDecimal.valueOf(v[i] - v[i - 1]).scaleByPowerOfTen(-prec);
    return result;
}

此方法使用Random.nextInt()均匀分布的属性.排序后,v[]数组的值是分割整个数量的点,因此您可以使用相邻元素之间的差异生成结果:

[   2,    5,   10,   11, ...,  197,  200]  // v[]
[0.02, 0.03, 0.05, 0.01, ...,  ..., 0.03]  // result[]

在这里,您使用整数值进行操作,因此舍入问题不再困扰.