IEMCO02-Editorial

Practice

DIFFICULTY:

EASY

PREREQUISITES: Data Structure, Strings

PROBLEM:

Judging by the current pandemic situation and classes being held in online mode, Prof. Ghosh decided to give an assignment to her students to check their knowledge regarding COVID 19 virus. The assignment was to enlist the components visible in the genome structure of Coronavirus. There are 500 students in the course and therefore checking 500 digital copies is tiresome. Prof. Ghosh wants you to develop a program for her using which she can check the copies fast and evaluate with ease.
The answer must consist of the given components :

  1. ACov(A)
  2. BCov (B)
  3. CCov©
  4. DCov (D)
  5. ECov (E)
  6. FCov (F)

Marking system:

All 6 components need to be enlisted for full marks.

The annotations attached to each component within braces need to be mentioned.

Each component name has 0.5 marks and the annotation carries 0.5 marks.

If a component is repeated still marks will be awarded only once for that component

The order in which the components are written in the answer does not matter.

The program if finds all the above-listed components in the answer of a student will award him or her 6 out of 6. If there is any spelling mistake found that component will not be considered as a valid answer and the corresponding marks will not be awarded. If the component name is correct but annotation is not provided or vice-versa then marks for only the part mentioned in the answer will be awarded.

Input : The components in the answer given by the student will be provided, each component in a single line.

Output : Depending on the marking system as per the number of correct answers given accordingly total marks scored will be generated. The output is the sum total marks scored by the student tallying with the answer submitted by him/her.

Constraints : At least one component should be mentioned and at most 6 components need to be mentioned.

Sample Input 1:

ACov(A)

BCov (B)

CCov©

DCov (D)

ECov (E)

FCov (F)

Output 1: 6 out of 6

EXPLANATION:

Explanation 1:

The input has 6 lines which indicate the student has written name of all 6 components needed. Now the program checks for spelling mistake or missing part of the answer in all 6 lines. In this example no error is found and therefore full marks is awarded i.e 6 out of 6.

Sample input 2 :

ACov(A)

BCov (B)

CCov

DCov (D)

ECov (E)

FCov

Sample output 2 : 5 out of 6

Explanation :

The input has 6 lines. So the student has written down all 6 components. But the program has to check whether the components have been written correctly complying to the guidelines or not. 2 annotations are missing in line 3 and line 6, therefore (0.5*2=1) is subtracted from full marks 6 and we get (6-1)=5 being awarded to the student.

Sample input 3:

ACov(A)

BCov (B)

CCoov©

DCov (D)

ECov (E)

FCov (F)

Sample output 3 : 5.5 out of 6

Explanation :

The input has 6 lines. Therefore the student has written all 6 components. But the program has to check whether all the components have been written correctly or not. In line no 3 there is a one spelling mistake so 0.5 marks for that component have been deducted from full marks 6 and the student is awarded 5.5 out of 6.

Test Case:

1

Input :

ACov(A)

BCov (B)

CCov©

DCov (D)

ECov (E)

FCov (F)

Output:

6 out of 6

2

Input :

ACov(A)

BCov (B)

CCov©

ECov (E)

FCov (F)

Output:

5 out of 6

3

Input :

ACov(A)

BCov

CCov©

DCov (D)

ECov (E)

FCov (F)

Output:

5.5 out of 6

4

Input :

ACov

BCov (B)

CCov

DCov (D)
DCov (D)

ECov (E)

OUTPUT

4 out of 6

5

Input :

DCov (D)

ECov (E)

FCov (F)

ACov(A)

BCov (B)

CCov©

Output:

6 out of 6

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