如何生成随机数，加起来就是一个明确的总数组？

问题描述

我需要随机生成与7个插槽Java中的数组。所有这些槽必须具有至少1的数值，但结合，具有的另一个限定数的合计值。他们也都需要有一个int值，没有1.5或0.9816465684646号码。例如：

I need to randomly generate an array with 7 slots in Java. All these slots must have a value of at LEAST 1, but combined, have a total value of another defined number. They also all need to be an int value, no 1.5 or 0.9816465684646 numbers.Example:

int a=10;

int[] ar = new int[7]
ar[0] = 1
ar[1] = 1
ar[2] = 2
ar[3] = 2
ar[4] = 1
ar[5] = 2
ar[6] = 1

我希望它产生类似的东西，但如果INT A = 15，所有数字将共有15个以任何顺序

I want it to generate something like that, but if int a=15, all the numbers would total 15 in any order

推荐答案

标准的方式来产生N个随机数的增加给定的准则是，想你之和为数字线，生成N-1个随机点行，排序他们，然后使用你的最终值点之间的差异。为了获得最小的1，从您的总和减去n启动，运行给出的算法，再加入1回各段。

The standard way to generate N random numbers that add to a given sum is to think of your sum as a number line, generate N-1 random points on the line, sort them, then use the differences between the points as your final values. To get the minimum 1, start by subtracting N from your sum, run the algorithm given, then add 1 back to each segment.

public class Rand {
    public static void main(String[] args) {
        int count = 8;
        int sum = 100;
        java.util.Random g = new java.util.Random();

        int vals[] = new int[count];
        sum -= count;

        for (int i = 0; i < count-1; ++i) {
            vals[i] = g.nextInt(sum);
        }
        vals[count-1] = sum;

        java.util.Arrays.sort(vals);
        for (int i = count-1; i > 0; --i) {
            vals[i] -= vals[i-1];
        }
        for (int i = 0; i < count; ++i) { ++vals[i]; }

        for (int i = 0; i < count; ++i) {
            System.out.printf("%4d", vals[i]);
        }
        System.out.printf("\n");
    }
}

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