如何四舍五入到指定数字的最接近倍数？

本文介绍了如何四舍五入到指定数字的最接近倍数？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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;