How you can Spherical as much as the Subsequent A number of of 5 in Python

By
Sohaib
-
0
10
Adv1


Adv2

The problem

Given an integer as enter, are you able to spherical it to the subsequent (which means, “larger”) a number of of 5?

Examples:

enter:    output:
0    ->   0
2    ->   5
3    ->   5
12   ->   15
21   ->   25
30   ->   30
-2   ->   0
-5   ->   -5
and so forth.

Enter could also be any constructive or damaging integer (together with 0).

You’ll be able to assume that every one inputs are legitimate integers.

The answer in Python code

Possibility 1:

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

Possibility 2:

def round_to_next5(n):
    whereas npercent5!=0:
        n+=1
    return n

Possibility 3:

import math
def round_to_next5(n):
    return math.ceil(n/5.0) * 5

Check circumstances to validate our resolution

inp = 0
out = round_to_next5(inp)
check.assert_equals(out, 0, "Enter: {}".format(inp))

inp = 1
out = round_to_next5(inp)
check.assert_equals(out, 5, "Enter: {}".format(inp))

inp = -1
out = round_to_next5(inp)
check.assert_equals(out, 0, "Enter: {}".format(inp))

inp = 5
out = round_to_next5(inp)
check.assert_equals(out, 5, "Enter: {}".format(inp))

inp = 7
out = round_to_next5(inp)
check.assert_equals(out, 10, "Enter: {}".format(inp))

inp = 20
out = round_to_next5(inp)
check.assert_equals(out, 20, "Enter: {}".format(inp))

inp = 39
out = round_to_next5(inp)
check.assert_equals(out, 40, "Enter: {}".format(inp))
Adv3