C ++随机数生成 [英] C++ random number generation

问题描述

例如，以下表达式:

r = (rand() % 10)+1;

从1-10生成一个随机数.

如何使它从0-10生成随机数?

How can we make it generate random numbers from 0-10?

谢谢.

推荐答案

您快到了！ rand()函数返回一个相当大范围(0到RAND_MAX)的随机值.使用模数运算符可以将此范围缩小到一个较小的范围(从0到9，因为您要乘以10)，然后+1将此范围从1移到10.

You're almost there! The rand() function returns a random value in a pretty large range (0 to RAND_MAX). The use of the modulus operator wraps this down into a smaller range (from 0 to 9, since you're modding by 10) and then the +1 moves this to be from 1 to 10.

要获取0到10之间的值，可以使用rand并将其值修改为11:

To get values between 0 and 10, you can take rand and mod its value by 11:

r = rand() % 11;

通常，要获得[0，n]范围内的随机值，您可以编写

More generally, to get random values in the range [0, n], you can write

r = rand() % (n + 1);

最后，要获得[k，n + k]范围内的值，您可以编写

And finally, to get values in the range [k, n + k], you can write

r = rand() % (n + 1) + k;

当然，正如纯粹主义者指出的那样，这不一定会给您真正统一的值，因为将rand()修改为某个值将不会平均分配所有整数.通常这不是问题(您将获得非常非常小的数量)，但是如果是这样，则可能需要考虑使用一个比rand()更强大的随机数生成器.

Of course, as purists will point out, this isn't necessarily going to give you truly uniform values because modding rand() by some value will not distribute all the integers evenly. This usually isn't a problem (you'll be off by a very, very small amount), but if it is you may want to consider looking into a more robust random number generator than rand().

这篇关于C ++随机数生成的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！