C ++：向上舍入到最接近的数字的倍数

Question

好的 - 我几乎很尴尬在这里张贴这个（如果有人投票结束我会删除），因为这似乎是一个基本问题。

这是在C ++中舍入到数字的倍数的正确方法吗？

我知道还有其他与此相关的问题，但我特别感兴趣的是知道在C ++中执行此操作的最佳方法是什么：

int roundUp(int numToRound, int multiple)

{

 if(multiple == 0)

 {

  return numToRound;

 }

 int roundDown = ( (int) (numToRound) / multiple) * multiple;

 int roundUp = roundDown + multiple; 

 int roundCalc = roundUp;

 return (roundCalc);

}

更新：对不起我可能没有明确意图。这里有些例子：

roundUp(7, 100)

//return 100

roundUp(117, 100)

//return 200

roundUp(477, 100)

//return 500

roundUp(1077, 100)

//return 1100

roundUp(52, 20)

//return 60

roundUp(74, 30)

//return 90

Answer 1

这适用于正数，不确定负数。它只使用整数数学。int roundUp(int numToRound, int multiple){&nbsp; &nbsp; if (multiple == 0)&nbsp; &nbsp; &nbsp; &nbsp; return numToRound;&nbsp; &nbsp; int remainder = numToRound % multiple;&nbsp; &nbsp; if (remainder == 0)&nbsp; &nbsp; &nbsp; &nbsp; return numToRound;&nbsp; &nbsp; return numToRound + multiple - remainder;}编辑：这是一个使用负数的版本，如果用“向上”表示结果总是> =输入。int roundUp(int numToRound, int multiple){&nbsp; &nbsp; if (multiple == 0)&nbsp; &nbsp; &nbsp; &nbsp; return numToRound;&nbsp; &nbsp; int remainder = abs(numToRound) % multiple;&nbsp; &nbsp; if (remainder == 0)&nbsp; &nbsp; &nbsp; &nbsp; return numToRound;&nbsp; &nbsp; if (numToRound < 0)&nbsp; &nbsp; &nbsp; &nbsp; return -(abs(numToRound) - remainder);&nbsp; &nbsp; else&nbsp; &nbsp; &nbsp; &nbsp; return numToRound + multiple - remainder;}

Answer 2

没有条件：int roundUp(int numToRound, int multiple)&nbsp;{&nbsp; &nbsp; assert(multiple);&nbsp; &nbsp; return ((numToRound + multiple - 1) / multiple) * multiple;}对于负数，这类似于从零开始舍入编辑：适用于负数的版本int roundUp(int numToRound, int multiple)&nbsp;{&nbsp; &nbsp; assert(multiple);&nbsp; &nbsp; int isPositive = (int)(numToRound >= 0);&nbsp; &nbsp; return ((numToRound + isPositive * (multiple - 1)) / multiple) * multiple;}如果multiple是2的幂int roundUp(int numToRound, int multiple)&nbsp;{&nbsp; &nbsp; assert(multiple && ((multiple & (multiple - 1)) == 0));&nbsp; &nbsp; return (numToRound + multiple - 1) & -multiple;}

Answer 3

当因子始终为正时，这种方法有效：int round_up(int num, int factor){&nbsp; &nbsp; return num + factor - 1 - (num - 1) % factor;}编辑：返回round_up(0,100)=100。请参阅下面的保罗评论，以获得返回的解决方案round_up(0,100)=0。