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))
```

## Answer 1, Authority 100%

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

## Answer 2

```
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])
```