algorithm - 如何找到给定范围内以大多数零十进制数字结尾的浮点值？

Question

有没有人有算法可以找到“min”和“max”之间的浮点值，当以十进制打印时以最多零位结尾(或者具有最少的小数位，换句话说)。当然可能有几种方案:在500到2500之间，1000或者2000都可以。就我的目的而言，两者都可以。 这是为了在 sndfile-spectrogram 中重新实现轴标记代码，所以我的目标语言是 C，但数学/伪代码很好。

Answer 1

如果我正确理解你的问题，我认为这将获得最高的数字。

float max = 432.334, min = 431.214; // The numbers you are looking between
float r = 0.01; // The first decimal place to check
float x0 = max, x1 = max; // Working variables
while (x1 > min)
{
    x0 = x1;
    x1 = r*floor(x1/r)
    r *= 10;
}

你的答案是 x0。

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编辑:更多解释

这是通过将较大的数字向下舍入为 10 的连续递增幂来实现的。r*floor(x1/r) 进行向下舍入。作为一个有效的例子:

r = 0.001
min, max = 0.1212, 0.1315
x1 = max

# Store the old value
x0 = x1
   = 0.1315

# Round down to the nearest r
x1 = r*floor(x1/r)
   = 0.001*floor(0.1315/0.001)
   = 0.001*floor(131.5)
   = 0.001*131
   = 0.131

# x1 is still larger than min, so multiply r by 10 and repeat
r  = 10*r
   = 10*0.001
   = 0.01
x0 = x1
   = 0.131
x1 = r*floor(x1/r)
   = 0.01*floor(0.131/0.01)
   = 0.01*floor(13.1)
   = 0.01*13
   = 0.13

# and again...
r  = 10*r
   = 10*0.01
   = 0.1
x0 = x1
   = 0.13
x1 = r*floor(x1/r)
   = 0.1*floor(0.13/0.1)
   = 0.1*floor(1.3)
   = 0.1*1
   = 0.1

# x1 is now smaller than min, so the loop ends. x0 is the last rounded value
# larger than min, so this is the answer.