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  1. Practice
  2. Algorithms
  3. Implementation
  4. Grading Students
  5. Discussions

Grading Students

  • Ardillo + 5 comments

    Python3 one-liner:

    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]

    • h201551072 + 2 comments

      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))
      • immanoj16 + 1 comment
        def solve(grades):
    result = []
    for i in grades:
        if i >= 38:
            if i % 5 >= 3:
                i += (5 - i % 5)
        result.append(i)
    return result
        • JehanJoshi + 0 comments

          You can club both the if conditions using an and here and neatly get it in one line :)

      • Mandar_Inamdar + 1 comment

        you are doing y=grades[i]/5 which will give you decimal numbers. round division using math functions or simply like: y=int(grades[i]/5)

        • ditonikola + 1 comment
          #!/bin/python3

import sys

def solve(grades):
    for k in range(0,n):
        if grades[k]<38:
            print(grades[k])
        else:
            l=grades[k]/10
            m=round(l)
            if 0.25<abs(l-m)<0.5 or 0.5<abs(l-m)<0.75:
                a=round(grades[k]/5)
                b=a*5
                c=b-grades[k]
                if c<0:
                    if c<-3:
                        print(b)
                    else:                    
                        print(grades[k])
                else:
                    if c<3:
                        print(b)
                    else:
                        print(grades[k])
            else:                        
                f=m*10
                d=f-grades[k]
                if d<0:
                    if d<-3:
                        print(f)
                    else:                    
                        print(grades[k])
                else:
                    if d<3:
                        print(f)
                    else:
                        print(grades[k])
            

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

          whats wrong with this pls tell me, it gives me runtime error, but passes 5 times

          • _dugy + 0 comments

            You have to return array of ints from "solve" function.

    • archit_imsec10 + 1 comment

      And how about this?

      map(lambda x: 5*(1 + x//5) if (x > 37 and ((x%5) > 2)) else x, grades)
      • ahmettortumlu25 + 1 comment

        so good.

        • sidra_mbn + 0 comments

          Thank you

    • Spiznak + 0 comments

      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...

    • Dimple151997 + 1 comment

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

      • MilosMM + 0 comments

        Hi, because we don't care about the value of the iterator.

    • rudralittle + 0 comments

      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]