C ++:舍入到数字的最接近的倍数 [英] C++: Rounding up to the nearest multiple of a number
问题描述
确定 - 我都不好意思在这里张贴这一点(我会删除,如果有人票关闭),因为它似乎是一个基本的问题
。这是围捕到一些在C ++中多了正确的方法是什么?
我知道有与此相关的其他问题,但我specficially兴趣知道什么是做这在C ++的最好方式:
INT综合报告(INT numToRound,诠释多)
{
如果(多个== 0)
{
返回numToRound;
}
INT ROUNDDOWN =((INT)(numToRound)/多)*倍数;
INT综述= ROUNDDOWN +多;
INT roundCalc =整;
返回(roundCalc);
}
更新: 对不起,我可能没做意图明显。下面是一些例子:
综合报告(7,100)
//返回100
综合报告(117,100)
//返回200
综合报告(477,100)
//返回500
综合报告(1077,100)
//返回1100
综合报告(52,20)
//返回60
综合报告(74,30)
//返回90
编辑:感谢所有的答复。以下是我去了:
INT综合报告(INT numToRound,诠释多)
{
如果(多个== 0)
{
返回numToRound;
}
INT余数= numToRound%多;
如果(余== 0)
{
返回numToRound;
}
返回numToRound +多 - 其余部分;
}
这适用于正数,不知道负面的。它仅使用整数运算。
INT综合报告(INT numToRound,诠释多)
{
如果(多个== 0)
返回numToRound;
INT余数= numToRound%多;
如果(余== 0)
返回numToRound;
返回numToRound +多 - 其余部分;
}
编辑:这是一个与负数的作品,版本,如果按向上你的意思是一个结果总是> =输入
。 INT综合报告(INT numToRound,诠释多)
{
如果(多个== 0)
返回numToRound;
INT余数= ABS(numToRound)%多;
如果(余== 0)
返回numToRound;
如果(numToRound℃,)
返回 - (ABS(numToRound) - 余数);
其他
返回numToRound +多 - 其余部分;
}
OK - I'm almost embarrassed posting this here (and I will delete if anyone votes to close) as it seems like a basic question.
Is this the correct way to round up to a multiple of a number in C++?
I know there are other questions related to this but I am specficially interested to know what is the best way to do this in C++:
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);
}
Update: Sorry I probably didn't make intention clear. Here are some examples:
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
EDIT: Thanks for all the replies. Here is what I went for:
int roundUp(int numToRound, int multiple)
{
if(multiple == 0)
{
return numToRound;
}
int remainder = numToRound % multiple;
if (remainder == 0)
{
return numToRound;
}
return numToRound + multiple - remainder;
}
This works for positive numbers, not sure about negative. It only uses integer math.
int roundUp(int numToRound, int multiple)
{
if (multiple == 0)
return numToRound;
int remainder = numToRound % multiple;
if (remainder == 0)
return numToRound;
return numToRound + multiple - remainder;
}
Edit: Here's a version that works with negative numbers, if by "up" you mean a result that's always >= the input.
int roundUp(int numToRound, int multiple)
{
if (multiple == 0)
return numToRound;
int remainder = abs(numToRound) % multiple;
if (remainder == 0)
return numToRound;
if (numToRound < 0)
return -(abs(numToRound) - remainder);
else
return numToRound + multiple - remainder;
}
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