Different possible marks for n questions and negative marking️‍🔥

Q: What is Different possible marks for n questions and negative marking?

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Given the number of questions as , and marks for the correct answer as and marks for the incorrect answer. One can either attempt to solve the question in an examination and get either marks if the answer is right, or marks if the answer is wrong, or leave the question unattended and get 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#

// 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