[TOC]
## 2024xCS50 Week 1 - C

### Psets

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#### Mario
problem: print hash pyramid below(less-comfy problem)
"o": space
"#" : hashes
<p class = "text-center"># o o #</p>
<p class = "text-center"># # o o # #</p>
<p class = "text-center"># # # o o # # #</p>
<p class = "text-center"># # # # o o # # # #</p>
<p class = "text-center"># # # # # o o # # # # #</p>

1. User enter a number to define how tall is the pyramid.
2. Print the pyramid.

##### Thought:
1. Start from left half and right half pyramid.
A left/right pyramid is made of " "(space) and "#"(hash symbol). Figure out relation between numbers of spaces and hashes per layer, turn it into statements.

    For the right pyrammid:
    // In the nth layer, print n's hashes in nth layer

    For the left pyramid, print spaces first and followed by hashes:
    // In the nth layer, print (HEIGHT - n -1) number of spaces
    // print n number of hashes

2. Combine these two pyramids together.
Don't forget the two spaces between the pyramids, since we want to print exactly the same pattern shown above.
Write a full pesudocode in higher level cites and finish all the statementes (For more information about the process of designing peudocode, you may see more from the book-Code Complete: Steve McConnell, Ch.9)

Here's my attempt:

// Ask user: "How tall is pyramid? "
// get HEIGHT(total layers) using get_int() from cs50.lib

// Starts the printing from layer n = 0 to layer height
    // In nth layer, print number of (HEIGHT - n - 1)'s spaces 
    // In nth layer, print number of n's hashes

    // In nth layer, print 2 spaces as the gap between left and right pyramids

    // In nth layer, print number of n's hashes

    // print "\n" and move to the next printing line

-->Translate to higher level description:

// for(layer = 0 to HEIGHT)
    // for(number = 0 to (HEIGHT - height -1))
    // print a space
    // End of for the second for-loop
    
    // Initialze a variable j = 0;
    // print j's hashes, 
    // repeat process from the last line, while (j < layer + 1)
    // End of the do-while loop
    
    // print two spaces
    
    // Initialize a variable k = 0;
    // print k's hashes
    // repeat this process while(k < layer + 1)
    // End of the do-while loop

    // print "\n" and go to the next line
// End of the first for-loop

3. Inplemente it and using debug tool if it didn't work out first time.

#### Cash
tbd
#### Credit
In this problem, you'll need to inplement an algorithm to check if user's entering card number is valid. If it's valid, print out its card type: VASA, AMEX or MasterCard. If it's not valid, print out message: "INVALID".

##### Thought: There're three functions need to be finished, one is check_card(), one is check_length() and the other is checksum().

Let's start from the main function():

// Ask user the number of credit card by printing : "Number: "

// If check_sum() passed Luhn's algorithm, then check its card number length

    // If passed card number length test, check which card is should be
    // Else if didn't pass card number test, print out message: "INVALID"

// Else if didn't passed check_sum(), print out message: "INVALID"

And here's demonstration for check_sum() and check_card():

1. check_sum(): In check_sum(), you'll need to understand how Luhn's algorithm operates first.
* First, look up all digit's number.
* Second, multiply even digit's number by two.
* Third, add it them up. Notice that if result of number added is above 2 digits, seperate it into 2 number, and added them up again.
* 4th, add to other left number which wern't be multiplied by 2.
* Finally, get the sum's last digit. If it's zero, then it passed the Luhn's algorithm.

// Declare a variable checksum to record the last digit in final step
// Declare a variable n to record current digit number seraching through
// Declare a variable sum to calculate sum from step1 to step4

// for( index = 0 to cardLength)
    // [block1]-get n 
    //if(index == cardLength -1) 
        // The last digit n of card number is the remainder of 10 by card number
    // Else
        // n is equal to : (Number from index+1 to index) - (number of index), then devided by 10
    
        
    //[block2]-multiply even digit bt 2
    // if (n is at even digit position)
        // multipy n by 2
   
    
    // [block3]-added n up to calculate sum
    // if (n is more than 2 digits)
        // sum = (first digit of n) + (last digit of n)
    // else (n has only one digit)
            // sum = n + previus sum
// End of for-loop

// Calculate the last digit of sum, which is the variavle checksum

//return checksum

2. check_card():

// Declare each card's criteria in some arrays containing the two starting number of AMEX, MasterCard, VISA
// Declare a variable to record inputing number's first two digits number
// Declare a status variable to identify which kind of card it is

// for(checkIndex = 0 to elements number of AMEX array)
    // If (inputStartNum == checkIndex'th element in AMEX array)
        // set status variable to 1
    // End of if statement
// End of for-loop
        
        // and so as the other cards.....


// write switch cases according to the value of status variable 

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### Additional practice

#### 1. Debug
Understand error message
#### 2. Half
Data type practice
#### 3. Prime
Practice using conditional statement: for loop

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