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# rounding number to the nearest whole

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According to the rules of the number, they are rounded to the nearest integer, and if the fractional part of the number is `0.5 `, the number is rounded up.

decided in this way, received `100/100 `, but I would like to get rid of the number `1000000 `, since when entering the number X, `4999999 `rounds Up:

``````from math import *
X = Float (Input ())
a = Floor (x)
b = x - a
if (int (Round (B, 1000000) * 100)) & lt; 50:
Print (int (x))
ELIF (INT (Round (B, 1000000) * 100)) == 50 OR (INT (ROUND (B, 2) * 100)) & gt; 50:
Print (Ceil (X))
``````

If you need to round the numbers in a specific side, then use the following functions

And if you need when a certain residue rounded the number in one direction or another, you can try to use the balance of the division (%) of the two numbers and compare already on it

`math.ceil () `– rounding numbers to the biggest

`math.floor () `– rounding numbers to the smaller side

`math.trunc () `– discard fractional part

source

``````def proper_round (num, dec = 0):
Num = STR (NUM) [: STR (NUM) .INDEX ('.') + Dec + 2]
IF num [-1] & gt; = '5':
RETURN FLOAT (NUM [: - 2- (NOT DEC)] + STR (int (NUM [-2- (NOT DEC)) + 1))
RETURN FLOAT (NUM [: - 1])
``````

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