3.5 C++ Accelerated - Koenig&Moo

//Exercise 3.5
//Create 2 vectors that can track student names and student grades
    //Assume that grades is a fixed number
    //we will assume 3

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using std::cout;    using std::cin;     using std::endl;
using std::vector;  using std::string;  using std::sort;


int main(){

    const int homework_count = 3;
    string name;
    double grade;
    double midterm;
    double final;
    double homework;

    //make vectors
    vector<string> names;
    vector<double> homeworks;
    vector<double> midterms;
    vector<double> finals;

    cout << "Enter Student names.  To exit enter EOF: ";
    while(cin >> name){
        names.push_back(name);
        //get mid terms
        cout << "Input mid term grade for " << name << ": ";
        cin >> midterm;
        midterms.push_back(midterm);
        //get finals
        cout << "Input final grade for " << name << ": ";
        cin >> final;
        finals.push_back(final);
        //get homeworks
        cout << "Input " << homework_count << " homework grades for " << name << ": ";
        for (int i = 0; i < homework_count; i++){
            cin >> homework;
            homeworks.push_back(homework);
        }
    }

    //Now that we have all grades and names, we can compute each final grade for each student

    vector<double>::size_type student_size = names.size();
    //need to find the median value
    int mid = homework_count / 2;
    for(int i = 0; i < student_size; i++){

        int offset = i * homework_count;
        double median;
        //sort the homework grades for this student
        sort(homeworks.begin() + offset, homeworks.begin() + offset  + homework_count);

        if(homework_count % 2){
            median = homeworks[mid + offset];
        }else{
            median = (homeworks[mid + offset] + homeworks[mid + offset - 1]) / 2;
        }

        cout << names[i] << " median hws: " << median << " average : " << .2 * midterms[i] + .4 * finals[i] + .4 * median << endl;
    }
    return 0;
}