Grading Students - Hacker Rank Solution

At HackerLand University, a passing grade is any grade 40 points or higher on a 100 point scale. Sam is a professor at the university and likes to round each student’s grade according to the following rules:

• If the difference between the grade and the next higher multiple of 5 is less than 3, round to the next higher multiple of 5
• If the grade is less than 38, don’t bother as it’s still a failing grade

Automate the rounding process then round a list of grades and print the results.

Input Format

The first line contains a single integer denoting  (the number of students).
Each line  of the  subsequent lines contains a single integer, , denoting student 's grade.

Constraints

Output Format

For each  of the  grades, print the rounded grade on a new line.

Sample Input 0

4
73
67
38
33

Sample Output 0

75
67
40
33

Explanation 0

The first grade,  is two below the next higher multiple of , so it rounds to .
 is  points less than the next higher multiple of  so it doesn’t round.
, like , rounds up to next higher multiple of , or  in this case.
 is less than , so it does not round.

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Grading Students - Hacker Rank Solution

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Solution

As long as , using  allows us to determine when we need to round a grade. We can then either:

1. Use a while loop to increment  to the next multiple of .

Python 2

n = int(raw_input().strip())

for a0 in xrange(n):
    grade = int(raw_input().strip())

    if grade>=38 and grade%5>=3:
        while grade%5!=0:
            grade = grade + 1
    print grade

C++

#include <bits/stdc++.h>
using namespace std;

void solution() {
     int n, x;
     cin>>n;
     for(int i=0; i<n; i++){
        cin>>x;
        if(x>=38 and x%5>=3){
            while(x%5!=0){
               x++;
            }
        }
        cout<<x<<endl;
     }
}

int main () {
    solution();
    return 0;
}

2. Add the result of  to .

Python 2

n = int(raw_input().strip())

for a0 in xrange(n):
    grade = int(raw_input().strip())

    if grade >= 38 and grade % 5 >= 3:
        grade = grade + 5 - (grade % 5)
        
    print grade

Or, even more efficiently:

Python 3

n = int(input().strip())
for a0 in range(n):
    grade = int(input().strip())

    if grade >= 38:
        # Here, we are only ever calculating 'grade mod 5' once:
        mod5 = grade % 5
        
        if mod5 >= 3:
            grade = grade + (5 - mod5)
        
    print(grade)

Problem Setter's code:

C++

#include <bits/stdc++.h>
#include<assert.h>

using namespace std;

void solution() {

     int n, x;

     cin >> n;

     assert(n > 0 && n <= 60);

     for(int i = 0; i < n; i++)
     {
        cin >> x;
        assert(x >= 0 && x <= 100);
        if(x >= 38)
        {
            int y = x;
            while(1)
            {
               if(y % 5 == 0)
    break;
               y++;
            }
            if(y - x <= 2)
             x = y;
        }
        cout << x << endl;
     }
}

int main () {
    
        solution();

    return 0;
}

Problem Tester's code:

import java.util.*;

public class Solution {

    public static int getRoundedGrade(int grade) {
        if (grade >= 38) {
            int mod5 = grade % 5;
            if (mod5 > 2) {
                grade += 5 - mod5;
            }
        }

        return grade;
    }

    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();
            System.out.println(getRoundedGrade(grade));
        }
        in.close();
    }
}

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Grading Students - Hacker Rank Solution

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