这是我的C代码。如果我理解正确的话，应该是这样的;

#include <stdio.h>

int main(){
int number;

printf("Enter number: \n");
scanf("%d" , &number);

if(number%5 == 0)
    printf("It is multiple of 5\n");
else{
    while(number%5 != 0)
        number++;
  printf("%d\n",number);
  }
}

这也是四舍五入到最接近5的倍数而不是四舍五入;

#include <stdio.h>

int main(){
int number;

printf("Enter number: \n");
scanf("%d" , &number);

if(number%5 == 0)
    printf("It is multiple of 5\n");
else{
    while(number%5 != 0)
        if (number%5 < 3)
            number--;
        else
        number++;
  printf("nearest multiple of 5 is: %d\n",number);
  }
}

如果有人需要“财务四舍五入”(0.5位总是向上):

def myround(x, base=5):
    roundcontext = decimal.Context(rounding=decimal.ROUND_HALF_UP)
    decimal.setcontext(roundcontext)
    return int(base *float(decimal.Decimal(x/base).quantize(decimal.Decimal('0'))))

根据文档，其他舍入选项是:

ROUND_CEILING(朝向无限) ROUND_DOWN(趋近于零) ROUND_FLOOR(朝向-∞) ROUND_HALF_DOWN(当平局趋于0时最接近) ROUND_HALF_EVEN(最接近偶数) ROUND_HALF_UP(到最接近的平局从0开始) ROUND_UP(远离零) ROUND_05UP(如果四舍五入后的最后一位为0或5，则远离0;否则趋于零)

默认情况下，Python使用ROUND_HALF_EVEN，因为它有一些统计优势(四舍五入的结果没有偏见)。

Use:

>>> def round_to_nearest(n, m):
        r = n % m
        return n + m - r if r + r >= m else n - r

它不使用乘法，也不会从/转换为浮点数。

四舍五入到最接近10的倍数:

>>> for n in range(-21, 30, 3): print('{:3d}  =>  {:3d}'.format(n, round_to_nearest(n, 10)))
-21  =>  -20
-18  =>  -20
-15  =>  -10
-12  =>  -10
 -9  =>  -10
 -6  =>  -10
 -3  =>    0
  0  =>    0
  3  =>    0
  6  =>   10
  9  =>   10
 12  =>   10
 15  =>   20
 18  =>   20
 21  =>   20
 24  =>   20
 27  =>   30

如你所见，它对负数和正数都适用。平局(例如-15和15)总是向上四舍五入。

一个类似的例子，四舍五入到5的最接近倍数，证明它也表现为不同的“基数”:

>>> for n in range(-21, 30, 3): print('{:3d}  =>  {:3d}'.format(n, round_to_nearest(n, 5)))
-21  =>  -20
-18  =>  -20
-15  =>  -15
-12  =>  -10
 -9  =>  -10
 -6  =>   -5
 -3  =>   -5
  0  =>    0
  3  =>    5
  6  =>    5
  9  =>   10
 12  =>   10
 15  =>   15
 18  =>   20
 21  =>   20
 24  =>   25
 27  =>   25

的值加上0.5，可以“欺骗”int()使其舍入而不是舍入 传递给int()的数字。

对于整数和Python 3:

def divround_down(value, step):
    return value//step*step


def divround_up(value, step):
    return (value+step-1)//step*step

生产:

>>> [divround_down(x,5) for x in range(20)]
[0, 0, 0, 0, 0, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15]
>>> [divround_up(x,5) for x in range(20)]
[0, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15, 20, 20, 20, 20]

def round_to_next5(n):
    return n + (5 - n) % 5