Beecrowd 1040 Average 3 Solution in C, C++, Python

Beecrowd 1040 Average 3 Solution

Question

Introduction

Beecrowd is a well-known site among competitive programmers focused on the enhancement of their problem-solving capacity. Europe, with all its resources, helps programmers perfect various algorithms and coding strategies typical to coding contests and interviews. Many users face problems that fall under this category, for instance, Beecrowd 1040: Average 3. In this section, we will analyze the problem in detail, figure out how to solve it, and what code is necessary in order to achieve it.

Understanding Beecrowd 1040: Average 3

Without jumping straight in to the code, it is important first to know what exactly Beecrowd 1040 is all about. It requires the students to find out their average score from four grades and decide if they succeeded or failed, but it also includes some re-examination if necessary.

Problem Statement Overview

One appropriate statement of the problem is exhibition 3 of Beecrowd 1040. In A496 the student has to construct several parameters based on four different examination scores for the student only.

Input and Output Requirements

Input: There are four floating point numbers each for every student’s score.

Output: It is the average as computed and the status of the student and if retaking the exam was needed it results.

Problem Deconstruction

Clarification of the Grading System

A four-grade system is adopted. They include N1, N2, N3, and N4. Each of the grades has its own weight:

N1 has a weight of 2

N2 has a weight of 3

N3 has a weight of 4

N4 has a weight of 1

How the Average is Calculated

Beecrowd 1040 Average 3 Solution in C

#include <stdio.h>
int main()
{
    float a, b, c, d, e, i, j, k;
    double avg, avrg;
    scanf("%f %f %f %f", &a, &b, &c, &d);
    avg = ((a*2) + (b*3) + (c*4) + d) / 10;
    printf("Media: %.1lf\n", avg);
    if ( avg < 5.0 ){
        printf("Aluno reprovado.\n");
    }
    else if( avg >= 7.0 ){
        printf("Aluno aprovado.\n");
    }
    else{
        printf("Aluno em exame.\n");//Beecrowd  1040 Average 3 Solution
        scanf("%f",&e);
        printf("Nota do exame: %.1f\n",e);
        avrg = (avg+e) / 2;
        if ( avrg >= 5.0 ){
            printf("Aluno aprovado.\n");
        }
        else {printf ("Aluno reprovado.\n");
        }
        printf("Media final: %.1lf\n",avrg);
    }
    return 0;
}

Beecrowd 1040 Average 3 Solution in C++

#include <iostream>
#include <iomanip>
using namespace std;

int main() {
    float a, b, c, d, e;
    double avg, avrg;

    cin >> a >> b >> c >> d;
    avg = ((a * 2) + (b * 3) + (c * 4) + d) / 10;
    cout << fixed << setprecision(1);
    cout << "Media: " << avg << endl;

    if (avg < 5.0) {
        cout << "Aluno reprovado." << endl;
    } else if (avg >= 7.0) {
        cout << "Aluno aprovado." << endl;
    } else {
        cout << "Aluno em exame." << endl;
        cin >> e;
        cout << "Nota do exame: " << e << endl;
        avrg = (avg + e) / 2;
        if (avrg >= 5.0) {
            cout << "Aluno aprovado." << endl;
        } else {
            cout << "Aluno reprovado." << endl;
        }
        cout << "Media final: " << avrg << endl;
    }

    return 0;
}

Beecrowd 1040 Average 3 Solution in Python

a, b, c, d = map(float, input().split())
avg = ((a * 2) + (b * 3) + (c * 4) + d) / 10
print(f"Media: {avg:.1f}")


if avg < 5.0:
    print("Aluno reprovado.")
elif avg >= 7.0:
    print("Aluno aprovado.")
else:
    print("Aluno em exame.")
    e = float(input())
    print(f"Nota do exame: {e:.1f}")
    avrg = (avg + e) / 2
    if avrg >= 5.0:
        print("Aluno aprovado.")
    else:
        print("Aluno reprovado.")
    print(f"Media final: {avrg:.1f}")

Wrap Up

Beecrowd 1040: Average 3 solution is excellent practice on weighted averages, if statements, and floating point numbers. In the present guide, you should be able to gradually work out the solution to the problem provided and you will be able to produce correct and optimal code.

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