Question

generate(vec.begin(), vec.end(), [=](){return static_cast<T>(static_cast<double>(ran())
/RAND_MAX*(max-min)+min); });

问题: RAND_max-min;

因此,我知道算法、手提法表述,并在此内容中包含条款。我的问题是完全如此。以上大胆的案文意味着什么。我要说的是,我知道自己是随机产生价值进程的一部分。但是,没有确切知道损失发生在哪里。因此,有人会pl倒这种微小的法典。

Answer 1

static_cast<double>ran()/RAND_ Fair*(max-min)+min”;

I m assuming you mistyped rand(), which returns a pseudorandom integer from 0 to RAND_MAX. Let s rewrite that in a way that clarifies the precedence a bit:

(T)(((双重) rand()/ RAND_ Fair) *(最大)+ min

因此:

rand(): take a random integer between 0 and RAND_MAX
(double) / RAND_MAX: divide as double by RAND_MAX, yielding a uniformly distributed double between 0 and 1:
* (max-min): multiply by the range (max-min), yielding a double from 0 to (max-min)
+min: add the minimum to yield a random double between min and max
static_cast<T>: cast it back to the original type

结果是在<代码>min和max之间统一分发的T类随机编号。

Answer 2

The expression static_cast<double>rand()/RAND_MAX creates a number between 0.0 and 1.0
When you multiply it by (max-min), you change the range, and when you add to min you shift the range. So, after that expression you have a random number (double) that ranges from min to max.

Answer 3

它的随机功能限于下游(因为地块可以归还零),并限于最大程度,因为即使它归还了100%的顶峰,并且添加到顶点上,你也处于顶峰。

Answer 4

You ll need to look at the entire expression: static_cast<double>(rand()) /RAND_MAX*(max-min)+min). Which with explicit grouping looks like: (static_cast<double>(rand()) /RAND_MAX)*(max-min)+min).
The first group returns a random value between 0 & 1 since rand() returns a value in the range 0 to RAND_MAX. The second group translates the 0 to 1 range to a min to max range.

Answer 5

ran( ran)功能在min和max之间产生随机值?

有时,我们需要的是A和B之间的随机价值(性别和最大)。因此,我们可以为此调整结果。

a is double, so we use a static_cast! 
a = rand()      ; 0     <=   a   <= RAND_MAX
a = a/RAND_MAX  ; 0     <=   a   <= 1
a = B * a       ; 0     <=   a   <= B
a = min +a      ; 0+min <=   a   <= B+min

a) 最大,

B+min = max
B = max-min

其他手法

a = rand()/RAND_MAX*(max-min) + min

最小和最大之间随机数量。

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