/* This program demonstrates
 * 1. The simple if statement
 * 2. Nested if statements
 * 3. Indentation for readability
 *
 * 1.  The simple if...else statement.
 * The if ...else statement is a way to get C++ to execute different lines
 * of code depending on the truth value of a condition. If the condition
 * were true, C++ executes the statements in the if block and skips over
 * the else block. If the condition were false, C++ skips over the if
 * block and executes the statements in the else block.
 * Once the if..else statement is done, C++ continues with the first line
 * after the if...else statement.
 * Both "if" and "else" are keywords.
 * 1. Only the if part requires a condition. NO CONDITION FOR ELSE
 * 2. The else part is optional, while the if part is not. WE CANNOT WRITE
 *    AN ELSE BLOCK WITHOUT THE IF BLOCK
 *
 * 2. The Nested if statement
 * Since the contents of the if block (Code that is executed if the condition is true)
 * can consist of multiple statements enclosed in {}, we can write an if statement
 * inside the if block (or the else block). This process is called Nesting. The
 * following example shows one level of Nesting, but there is no theoretical limit to
 * the number of nested levels.
 * Also, we can have multiple if statements with nesting and nested if statements, where
 * the internal level is a  multiple if. Any combination of control structures is allowed
 * as long they are syntactically OK.
 *
 * 3. Style Guidelines - Indentation
 * It is standard practice to indent code to make it readable.
 * Whenever we begin a block of code by opening '{', we go one tab in, and whenever we end a
 * block of code with a '}', we go one tab out.
 * This gives us 3 advantages - we can easily match which '{' corresponds to which '}'
 * We can make sure there are no missing or stray '{' or '}', which would cause compiler errors.
 * We can easily understand which lines of code belong in a cod block.
 */
#include<iostream>
using namespace std;
int main()
{
    /*  Here, we calculate if a student will pass the class.
    * you can pass this class only if you have a C- on the entire class AND
    * a C- average on the milestones.
    * Here we read in the grades. Practice exercise, discussion sessions and class
    * participation on 8% each, exams on 39% and hw assignments on 35%
    */

    double cp,pe,ds,exam, hw;   //already weighted grades for class participation, practice
    // exercises, discussion sessions, exams and hw assignments
    cout<<"Enter the grades: ";
    cin>>cp>>pe>>ds>>exam>>hw;
    double mileAvg = exam / 39 * 100 ;      // test avgerage on 100
    double grade = cp + ds + pe + exam +hw + 2; // 2 points for 1st day quiz, help sessions

    //Now, we are ready to write the if statement
    //To pass, students need 70% overall and 70% average on milestone assignments.
    if (grade>=70 && mileAvg >=70)
    {
        cout<<"You passed! Congratulations!"<<endl;
    }
    else
    {
        cout<<"Please try again!"<<endl;
    }

    cout<<"Your grade was "<<grade<<endl;
    cout<<"Your test average was "<< mileAvg<<endl;

    //nested if statement
    /* Given 3 numbers, find the largest
     * 1. Input - read in 3 numbers - declare 3 variables, cin (x,y,z)
     * 2. Find the largest
     * 2.1 x>y? If yes, 2.1.1. If no, 2.2
     * 2.1.1 x>z? If yes, largest = x. Otherwise largest = z. Go to step 3
     * 2.2 y>z? If yes, largest =y. Otherwise largest = z.
     * 3. Print the largest - cout
     */

    int x,y,z;
    cout<<"Enter 3 distinct numbers: ";
    cin>>x>>y>>z;
    int largest;
    if (x > y)
    {
        if(x > z)
        {
            largest = x;
        }
        else {
            largest = z;
        }
    } else {
        if (y > z) {
            largest = y;
        }
        else{
            largest = z;
        }
    }
    // Note the indentation for the above structure. 
    
    cout<<largest<<" is the largest number"<<endl;
}