生成总和为常数的随机数 [英] generate random numbers of which the sum is constant

问题描述

我在想是否有办法生成一组总和始终为常数的随机数.例如，20 可以分为 5 个数字(1、2、3、4、10)我不在乎这 5 个数字中的每一个是什么，只要它们的总和等于 20.无论如何以编程方式这样做?

I am thinking if there is anyway to generate a set of random numbers of which the sum is always a constant. For example, 20 can be divided into 5 numbers ( 1, 2,3,4,10) I don't care what each of the 5 numbers is as long as their sum is equal to 20. Is there anyway to do so programmatically?

推荐答案

要获得均匀分布，诀窍是将总和视为数轴，而不是为段生成随机数，而是生成 n-1数字作为沿线的点，然后减去以获得线段.这是 ojrandlib 中的函数:

To get a uniform distribution, the trick is to think of your sum as a number line, and rather than generating random numbers for the segments, generate n-1 numbers as points along the line, and subtract to get the segments. Here's the function from ojrandlib:

static int compare(const void *a, const void *b) {
    return *(int*)a - *(int*)b;
}
void ojr_array_with_sum(ojr_generator *g, int *a, int count, int sum) {
    int i;
    for (i = 0; i < count-1; ++i) { a[i] = ojr_rand(g, sum+1); }
    qsort(a, count-1, sizeof(int), compare);
    a[count-1] = sum;
    for (i = count-1; i > 0; --i) { a[i] -= a[i-1]; }
}

ojr_rand(g, limit) 生成一个从 0 到 limit-1 的统一随机整数.这个函数然后用 count 个随机整数填充数组 a，这些整数与 sum 相加.将此适应任何其他 RNG 应该不会太难.

ojr_rand(g, limit) generates a uniform random integer from 0 to limit-1. This function then fills the array a with count random integers that add to sum. Shouldn't be too hard to adapt this to any other RNG.

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