如何四舍五入到指定数字的最接近倍数？

问题描述

我已经看过很多关于将数字四舍五入到最接近的倍数的问题，但是我对它们的方法不够了解，无法采用它们四舍五入到45，或者他们使用其他语言的特定语言方法。

I've looked at many questions regarding rounding up to the nearest multiple of a number, but I can't understand their methods well enough to adopt them for rounding up to 45 or they use language specific methods of other languages.

如果上述内容还没有多大意义，这里有一些更详细的解释：

Here is a bit more detailed explanation if the above doesn't make much sense yet:

示例输入：

int num1 = 67;
int num2 = 43;

预期结果：

>>round(num1) = 90
>>round(num2) = 45

推荐答案

足以添加缺少的 mod45 ：

int upperround45(int i) {
    int temp = i%45;
    //For the algorithm we wish the rest to be how much more than last multiple of 45.
    if (temp < 0 )
        temp = 45 + temp;
    if (temp == 0) {
        return i;
    } else {
        return i + 45 - temp;
    }
}

编辑：

通常：

int upperround(int num, int base) {
    int temp = num%base;
    if (temp < 0 )
        temp = base + temp;
    if (temp == 0)
        return num;
    return num + base - temp;

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