java – 将给定数字M随机分成N个部分

发布时间：2020-12-15 04:29:06 所属栏目：Java 来源：网络整理

导读：因此,我所拥有的想法是能够将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位小数且小于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

因此,我所拥有的想法是能够将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位小数且小于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：

Getting N random numbers that the sum is M

解决方法

假设您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 []数组的值是整个金额被拆分的点,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[]
[0.02,0.03,0.03]  // result[]

在这里,您使用整数值进行操作,因此舍入问题不再困扰.

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!