Different possible marks for n questions and negative marking

Last Updated : 08 Sep, 2022

Given the number of questions as n , and marks for the correct answer as p and q marks for the incorrect answer. One can either attempt to solve the question in an examination and get either p marks if the answer is right, or q marks if the answer is wrong, or leave the question unattended and get 0 marks. The task is to find the count of all the different possible marks that one can score in the examination.

Examples:

Input: n = 2, p = 1, q = -1
Output: 5
The different possible marks are: -2, -1, 0, 1, 2

Input: n = 4, p = 2, q = -1
Output: 12

Approach: Iterate through all the possible number of correctly solved and unsolved problems. Store the scores in a set containing distinct elements keeping in mind that there is a positive number of incorrectly solved problems.

Below is the implementation of the above approach:

C++

// CPP program to find the count of
// all the different possible marks
// that one can score in the examination
#include<bits/stdc++.h>

using namespace std;

    // Function to return
    // the count of distinct scores
    int scores(int n, int p, int q)
    {
        // Set to store distinct values
        set<int> hset;

        // iterate through all
        // possible pairs of (p, q)
        for (int i = 0; i <= n; i++) 
        {
            for (int j = 0; j <= n; j++) 
            {

                int correct = i;
                int not_solved = j;
                int incorrect = n - i - j;

                // if there are positive number
                // of incorrectly solved problems
                if (incorrect >= 0)
                    hset.insert(p * correct
                            + q * incorrect);
                else
                    break;
            }
        }

        // return the size of the set
        // containing distinct elements
        return hset.size();
    }

    // Driver code
    int main()
    {

        // Get the number of questions
        int n = 4;

        // Get the marks for correct answer
        int p = 2;

        // Get the marks for incorrect answer
        int q = -1;

        // Get the count and print it
        cout << (scores(n, p, q));
    }

// This code is contributed by
// Surendra_Gangwar

Java

// Java program to find the count of
// all the different possible marks
// that one can score in the examination

import java.util.*;

class GFG {

    // Function to return
    // the count of distinct scores
    static int scores(int n, int p, int q)
    {
        // Set to store distinct values
        HashSet<Integer>
            hset = new HashSet<Integer>();

        // iterate through all
        // possible pairs of (p, q)
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= n; j++) {

                int correct = i;
                int not_solved = j;
                int incorrect = n - i - j;

                // if there are positive number
                // of incorrectly solved problems
                if (incorrect >= 0)
                    hset.add(p * correct
                             + q * incorrect);
                else
                    break;
            }
        }

        // return the size of the set
        // containing distinct elements
        return hset.size();
    }

    // Driver code
    public static void main(String[] args)
    {

        // Get the number of questions
        int n = 4;

        // Get the marks for correct answer
        int p = 2;

        // Get the marks for incorrect answer
        int q = -1;

        // Get the count and print it
        System.out.println(scores(n, p, q));
    }
}

Python3

# Python3 program to find the count of 
# all the different possible marks 
# that one can score in the examination 

# Function to return the count of
# distinct scores 
def scores(n, p, q): 
    
    # Set to store distinct values 
    hset = set() 

    # Iterate through all possible
    # pairs of (p, q) 
    for i in range(0, n + 1): 
        for j in range(0, n + 1): 

            correct = i 
            not_solved = j 
            incorrect = n - i - j 

            # If there are positive number 
            # of incorrectly solved problems 
            if incorrect >= 0:
                hset.add(p * correct +
                         q * incorrect) 
            else:
                break

    # return the size of the set 
    # containing distinct elements 
    return len(hset) 
    
# Driver code 
if __name__ == "__main__":
    
    # Get the number of questions 
    n = 4

    # Get the marks for correct answer 
    p = 2

    # Get the marks for incorrect answer 
    q = -1

    # Get the count and print it 
    print(scores(n, p, q)) 
    
# This code is contributed by Rituraj Jain

// C# program to find the count of
// all the different possible marks
// that one can score in the examination
using System;
using System.Collections.Generic; 

class GFG 
{

    // Function to return
    // the count of distinct scores
    static int scores(int n, int p, int q)
    {
        // Set to store distinct values
        HashSet<int>
            hset = new HashSet<int>();

        // iterate through all
        // possible pairs of (p, q)
        for (int i = 0; i <= n; i++) 
        {
            for (int j = 0; j <= n; j++) 
            {

                int correct = i;
                int not_solved = j;
                int incorrect = n - i - j;

                // if there are positive number
                // of incorrectly solved problems
                if (incorrect >= 0)
                    hset.Add(p * correct
                            + q * incorrect);
                else
                    break;
            }
        }

        // return the size of the set
        // containing distinct elements
        return hset.Count;
    }

    // Driver code
    public static void Main()
    {

        // Get the number of questions
        int n = 4;

        // Get the marks for correct answer
        int p = 2;

        // Get the marks for incorrect answer
        int q = -1;

        // Get the count and print it
        Console.WriteLine(scores(n, p, q));
    }
}

/* This code contributed by PrinciRaj1992 */

JavaScript

<script>

// JavaScript program to find the count of
// all the different possible marks
// that one can score in the examination

    // Function to return
    // the count of distinct scores
    function scores(n, p, q)
    {
        // Set to store distinct values
        let hset = new Set();

        // iterate through all
        // possible pairs of (p, q)
        for (let i = 0; i <= n; i++) {
            for (let j = 0; j <= n; j++) {

                let correct = i;
                let not_solved = j;
                let incorrect = n - i - j;

                // if there are positive number
                // of incorrectly solved problems
                if (incorrect >= 0)
                    hset.add(p * correct
                             + q * incorrect);
                else
                    break;
            }
        }

        // return the size of the set
        // containing distinct elements
        return hset.size;
    }
    
// Driver Code

           // Get the number of questions
        let n = 4;

        // Get the marks for correct answer
        let p = 2;

        // Get the marks for incorrect answer
        let q = -1;

        // Get the count and print it
        document.write(scores(n, p, q));
        
</script>

Output

Find the total marks obtained according to given marking scheme

AmanKumarSingh

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Different possible marks for n questions and negative marking

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