如何在给定间隔中以相等的概率生成随机数 [英] How to generate a random number with equal probability in a given interval

问题描述

我做了很多尝试，但未能解决此问题

I tried a lot but could not get a solution for this problem

函数返回范围为的数字[ 1,6] 的概率相等。您可以使用库的 rand（）函数，并假设实现 rand（）返回范围内的数字 [0，RAND_MAX] 的概率均等。

Function returns numbers in range [1,6] with equal probability. You can use library's rand() function and you can assume implementation of rand() returns number in range number in range [0,RAND_MAX] with equal probability.

推荐答案

我们将分多个步骤进行。

We'll do this in multiple steps.

您需要生成一个范围为 [1、6] （含）。

You need to generate a number in the range [1, 6], inclusive.

您有一个随机数生成器，该生成器将生成范围 [0..RAND_MAX] 。

You have a random number generator that will generate numbers in the range [0..RAND_MAX].

假设您要生成 [0..5] 。您可以这样做：

Let's say you wanted to generate numbers in the range [0..5]. You can do this:

int r = rand();  // gives you a number from 0 to RAND_MAX
double d = r / RAND_MAX;  // gives you a number from 0 to 1
double val = d * 5; // gives you a number from 0 to 5
int result = round(d);  // rounds to an integer

您可以使用该技术对给定范围 [0，高] ，您可以生成一个随机数，除以 RAND_MAX ，再乘以 high ，然后对结果取整。

You can use that technique to So given a range of [0, high], you can generate a random number, divide by RAND_MAX, multiply by high, and round the result.

您的范围是 [1，6] ，因此您必须添加另一步骤。您要生成一个在 [0，5] 范围内的随机数，然后加1。或者，通常，要在给定范围内生成一个随机数， [低，高] ，您输入：

Your range is [1, 6], so you have to add another step. You want to generate a random number in the range [0, 5], and then add 1. Or, in general, to generate a random number in a given range, [low, high], you write:

int r = rand();
double d = r / RAND_MAX;
int range = high - low + 1;
double val = d * range;
result = round(val);

很明显，您可以合并其中的一些操作。我只是单独显示它们以进行说明。

Obviously you can combine some of those operations. I just showed them individually to illustrate.

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