对已接受答案的补充，用于指定四舍五入到最接近的5或其他值

import math

def my_round(x, base, down = True):
    return base * math.floor(x/base) + (not down) * base

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]

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);
  }
}

下一个5的倍数

考虑51需要转换为55:

code here

mark = 51;
r = 100 - mark;
a = r%5;
new_mark = mark + a;

round(x[， n]):数值四舍五入到10的负n次方的最接近倍数。所以如果n是负的…

def round5(x):
    return int(round(x*2, -1)) / 2

由于10 = 5 * 2，您可以对2使用整数除法和乘法，而不是对5.0使用浮点除法和乘法。这并不重要，除非你喜欢位移位

def round5(x):
    return int(round(x << 1, -1)) >> 1

divround的修改版本:-)

def divround(value, step, barrage):
    result, rest = divmod(value, step)
    return result*step if rest < barrage else (result+1)*step