是否有一个内置函数可以像下面这样舍入?
10 -> 10
12 -> 10
13 -> 15
14 -> 15
16 -> 15
18 -> 20
是否有一个内置函数可以像下面这样舍入?
10 -> 10
12 -> 10
13 -> 15
14 -> 15
16 -> 15
18 -> 20
当前回答
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);
}
}
其他回答
那么这个呢:
def divround(value, step):
return divmod(value, step)[0] * step
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;
我需要四舍五入到前面的5。
举个例子,16发到15发或者19发到15发
下面是使用的代码
def myround(x,segment):
preRound = x / segment
roundNum = int(preRound)
segVal = segment * roundNum
return segVal
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);
}
}
def round_up_to_base(x, base=10):
return x + (base - x) % base
def round_down_to_base(x, base=10):
return x - (x % base)
这给了
基础= 5:
>>> [i for i in range(20)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>> [round_down_to_base(x=i, base=5) for i in range(20)]
[0, 0, 0, 0, 0, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15]
>>> [round_up_to_base(x=i, base=5) for i in range(20)]
[0, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15, 20, 20, 20, 20]
基础= 10:
>>> [i for i in range(20)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>> [round_down_to_base(x=i, base=10) for i in range(20)]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10]
>>> [round_up_to_base(x=i, base=10) for i in range(20)]
[0, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 20, 20, 20, 20, 20, 20, 20, 20, 20]
在Python 3.7.9中测试