HackerLand University has the following grading policy:
- Every student receives a grade in the inclusive range from 0 to 100.
- Any grade less than 40 is a failing grade.
Sam is a professor at the university and likes to round each student’s grade according to these rules:
- 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.
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Examples
- grade = 84 round to 85 (85 – 84 is less than 3)
- grade = 29 do not round (result is less than 40)
- grade do not round (60 – 57 is 3 or higher)
Given the initial value of for each of Sam’s students, write code to automate the rounding process.
Function Description
Complete the function gradingStudents in the editor below.
gradingStudents has the following parameter(s):
- int grades[n]: the grades before rounding
Returns
- int[n]: the grades after rounding as appropriate
Input Format
The first line contains a single integer,n, the number of students.
Each line i of the n subsequent lines contains a single integer, grade[i].
Constraints
- 1<=n<=60
- 0<=grades[i]<=100
Sample Input 0
4 73 67 38 33
Sample Output 0
75 67 40 33
Grading Students HackerRank Solution in C
#include <math.h> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <assert.h> #include <limits.h> #include <stdbool.h> int main(){ int n; scanf("%d",&n); for(int a0 = 0; a0 < n; a0++){ int grade; scanf("%d",&grade); // your code goes here int x = (grade+4)/5; x *= 5; if(x>=40 && x-grade<3)grade=x; printf("%d\n",grade); } return 0; }
Grading Students HackerRank Solution in C++
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <string> #include <bitset> #include <cstdio> #include <limits> #include <vector> #include <climits> #include <cstring> #include <cstdlib> #include <fstream> #include <numeric> #include <sstream> #include <iostream> #include <algorithm> #include <unordered_map> using namespace std; int main(){ int n; cin >> n; for(int a0 = 0; a0 < n; a0++){ int grade; cin >> grade; if (grade >= 38) { int rem = grade % 5; if (rem >= 3) grade += 5 - rem; } cout << grade << endl; } return 0; }
Grading Students HackerRank Solution in Java
import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); for(int a0 = 0; a0 < n; a0++){ int grade = in.nextInt(); // your code goes here if(grade < 38){ System.out.println(grade); } else{ int q = grade/5; int rem = grade%5; if(rem >= 3){ System.out.println((q+1)*5); } else{ System.out.println(grade); } } } } }
Grading Students HackerRank Solution in Python
#!/bin/python import sys n = int(raw_input().strip()) for a0 in xrange(n): grade = int(raw_input().strip()) # your code goes here if grade < 38: print grade continue if grade % 5 >= 3: grade = ((grade / 5) + 1) * 5 print grade
Grading Students HackerRank Solution in C#
using System; using System.Collections.Generic; using System.IO; using System.Linq; class Solution { static void Main(String[] args) { int n = Convert.ToInt32(Console.ReadLine()); for(int a0 = 0; a0 < n; a0++){ int grade = Convert.ToInt32(Console.ReadLine()); int final = 0; int next = ((grade / 5) + 1) * 5; if (next - grade >= 3 || grade < 38) final = grade; else final = next; Console.WriteLine(final); } } }
Attempt – Grading Students HackerRank Challenge
Link – https://www.hackerrank.com/challenges/grading/
Next HackerRank Challenge Solution
Link – https://exploringbits.com/apple-and-oranges-hackerrank-solution/
Aayush Kumar Gupta is the founder and creator of ExploringBits, a website dedicated to providing useful content for people passionate about Engineering and Technology. Aayush has completed his Bachelor of Technology (Computer Science & Engineering) from 2018-2022. From July 2022, Aayush has been working as a full-time Devops Engineer.