c - 四舍五入到给定值的最接近倍数

Question

如果我们有一个任意double值f、另一个值v和一个乘法因子p，我怎样才能将该值捕捉f到最接近的v幂p？

例子：

• f = 3150.0
• v = 100.0
• p = 2

乘法会像这样

• 100 ( v)
• 200（乘以p）
• 400
• 800
• 1600
• 3200

...

f最接近3200.0所以函数应该返回3200.0

实际上有一个名字，我似乎忘记了，也许这就是我找不到这样一个功能的原因。

标签： cmath

Answer 1

Let k = floor(log_p(f/v)) where log_p(x) = log(x)/log(p) is the logarithm to base p function. It follows from the properties of floor and log that p^k v <= f < p^(k+1) v, which gives the two closest values to f of the form p^n v.

Which of those two values to choose depends on the exact definition of "nearest" in your use-case. If taken in the multiplicative sense (as would be natural on a log scale), that "nearest" value can be calculated directly as p^n v where n = round(log_p(f/v)) = round(log(f/v)/log(p)).