# Grading Students | HackerRank Solution

Hello coders, today we are going to solve Grading Students HackerRank Solution which is a Part of HackerRank Algorithm Series.

HackerLand University has the following grading policy:

• Every student receives a grade in the inclusive range from 0 to 100.

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.

Examples

• grade = 84 round to 85 (85 – 84 is less than 3)
• grade = 29 do not round (result is less than 40)
• grade = 57 do not round (60 – 57 is 3 or higher)

Given the initial value of grade for each of Sam’s n students, write code to automate the rounding process.

Function Description

Complete the function gradingStudents in the editor below.

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, grades[i].

## Constraints

• 1 <= n <= 60
• 0 <= grades[i] <= 100

Sample Input 0

``````4
73
67
38
33``````

Sample Output 0

``````75
67
40
33``````

Explanation 0

1. Student 1 received a 73, and the next multiple of 5 from 73 is 75. Since 75 – 73 < 3, the student’s grade is rounded to 75.
2. Student 2 received a 67, and the next multiple of 5 from 67 is 70. Since 70 – 67 = 3, the grade will not be modified and the student’s final grade is 67.
3. Student 3 received a 38, and the next multiple of 5 from 38 is 40. Since 40 – 38 < 3, the student’s grade will be rounded to 40.

### 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;
}```

### Python

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