Thanks. A lengthier solution might improve readability but compromise on lines of code (maintainability). Either way it doesn't affect runtime, so it comes down to a matter of personal preference I guess. :)

A simple and perhaps readable solution.. You can use n instead of temp as its just a dummy..

int main() 
{
    int n,temp;
    cin>>temp;
    while(cin>>n)
    {
        if(n<38)
            cout<<n<<endl;
        else if(n%5 >= 3)
            cout<<n+ (5-n%5)<<endl;
        else
            cout<<n<<endl;

    }
    return 0;
}

You can do the same thing in C# ->

return grades.Select(grade => grade < 38 || (grade % 5 < 3) ? grade :  grade + (5 - grade % 5 )).ToArray();

return grades.Select(grade => grade < 38 || (grade % 5 < 3) ? grade : grade + (5 - grade % 5 )).ToArray();

Don't know why I got downvoted for a simple comment?

nice shortcut for nearest 5th divisible value.

g+5 - g%5

Does not working out.It did some basics mistakes on the Editor.

The rules for rounding off is stated as follows :

If the difference between the grade and the next multiple of 5 is less than 3, round grade up to the next multiple of 5.

If the value of grade is less than 38, no rounding occurs as the result will still be a failing grade.

Since the value 67 is greater than 38 and the difference between value and next multiple of 5 (i.e. 70) is 3 (70 - 67) which is not less than 3, it will not be rounded. Hence, 67 i.e. 'grade' value is printed.

However, for the value 68, the difference is 2 (70 - 68) < 3, so the grade will be rounded to the next multiple by the calculation :

grade + (5 - (grade % 5)) = 68 + (5 - (68 % 5))

= 68 + (5 - 3)

= 68 + 2

= 70 (next multiple of 5)

Hope this helps! :)

Just a tip I've learned from my professors, you should rarely ever have to rewrite the same lines of code, see how your initial if and your default else are doing the same thing? You could condense this by doing this:

if(n>38 && n%5 > 3)
{
    cout << n+(5-n%5) << endl;
}
else
{
    cout << n << endl;
}

Makes it still readable, but in less lines of code. Also, in Industry code it's generally good practice to use curly braces on the body of if statements to prevent any confusion.

i am new in pyhton can u help me with the mistake?

#!/bin/python

import sys

def solve(grades):
    r=[]
    y=0
    d=0
    # Complete this function
    for i in range(0,len(grades)-1):
        if(grades[i]>=38):
            y=grades[i]/5
            d=5*(y+1)-grades[i]
            if(d<3):
                r[i]=5*(y+1)
            else:
                r[i]=grades[i]
        else:
            r[i]=grades[i]
    return r
               

n = int(raw_input().strip())
grades = []
grades_i = 0
for grades_i in xrange(n):
    grades_t = int(raw_input().strip())
    grades.append(grades_t)
result = solve(grades)
print "\n".join(map(str, result))

And how about this?

map(lambda x: 5*(1 + x//5) if (x > 37 and ((x%5) > 2)) else x, grades)

1 iteration, almost 1-liner. O(n) is slightly better than O(2n):

for _ in range(int(input())):
    x = int(input())
    print(x + (5 - x % 5)) if x >= 38 and 5 - x % 5 < 3 else print(x)

I wouldn't be surprised to learn that there is a better way to do it, though...

Can you explain the code.. why do we write __ in for loop and 2nd line in [] brackets

i ran it on my jupyer notebook. i cant understand the output. input>>>>>>>>>>>>> grades = [int(input()) for __ in range(int(input()))]

[print(g+5 - g%5 if g%5 > 2 and g>37 else g) for g in grades]

output>>>>>> 5 45 66 23 56 98 45 66 23 56 100 [None, None, None, None, None]

The problem states that the rounding should occur if the DIFFERENCE is LESS than 3. The difference here is 3 so it is not rounded.

haha I hope you're joking

awsome! your code!

i dont think so it pass all the test cases ...but thanks for your kindness

One liner for abouve code

def gradingStudents(grades): return [g if g < 38 or g % 5 < 3 else g + (5 -(g % 5)) for g in grades]

this solution is mathematically cool man

You are using grade, which is a list. You have to access the indices in the list using loop like this

for i in grade:   
    if i >= 38:
        if i % 5 == 3:
            i += 2
        elif i % 5 == 4:
            i += 1
    print (i)

fgfhyhyrihowhhflhiihiw{} bhai yeah galti hai teri program mmm

Which length of output you want to display ? which size of array would display? for all thoswe que solution is )*result_size = grades_size;