如何得到最小的数字A，即N倍于另一个数字B，以及更大或等于数字C

Question

我想得到一个数字（floatB）的N倍的最小数字（floatA），并且这个数字必须大于或等于另一个数字（floatC），例如

float_c = 20.0
float_b = 7.0
than number that is N times float_b is: 21, 28, 35 ...
so float_a is 21.0

像下面这样的一些代码可能不起作用（我不确定）

let tmp = float_c / float_b;
let tmp = tmp.ceil();
float_a = tmp * float_b;

/*
e.g: 
float_c = 21.0
float_b = 7.0;
let tmp = 21.0 /7.0;  // 3.00000000000001;
let tmp = tmp.ceil();   // 4.0
float_a = tmp * 7.0; //28, that's worng
*/

下面的代码可以工作吗？

float_c = 21.0
float_b = 7.0;
let tmp = 21.0 / 7.0; // 3.00000000001
let tmp2 = tmp.ceil(); // 4.0
let float_a = tmp2 * float_b;
let float_backtest = float_a - float_b;

if (float_backtest - float_c).abs() < epsilon:
    return float_c
else:
    return float_a;

Answer 1

如果

c

和

b

为正，则此 C 代码生成大于

b

的

c

的最小倍数：

// If c < b, b is the first multiple of b greater than or equal to c.
if (c < b)
    a = b;
else
{
    // Get the remainder of c modulo b.  This is exact by nature of fmod.
    double r = fmod(c, b);

    // If c is exactly a multiple of b, it is the desired result.
    if (r == 0)
        a = c;

    /* Otherwise, calculate the amount to add to c to make it a multiple,
       then add it. "b - r" is exact because r cannot have any bits lower
       in position value than b has (because b <= c) nor any higher
       (because r < b). The addition of c may be inexact, but then the
       desired result is not representable.
    */
    else
        a = b - r + c;
}