This activity was easy. No programming language required work, maybe that's why I feel it's easy to complete.

This commit is contained in:
Seoyeon Lee 2024-12-08 00:03:24 -05:00
parent 44f875c624
commit 122a47f7d4
1 changed files with 26 additions and 6 deletions

View File

@ -37,19 +37,39 @@ return digit_names[number]
## Integers under 20
If the integer is under 10, then use the procedure described above.
Otherwise, ... (this is where you take over!)
digit_names2 = [
"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen",
"eighteen", "nineteen"
]
to get the digit, we devide the number by 10, and consider the remainder.
return dignt_names2 [number%10]
## Integers under 100
If the integer is under 20, use the previous procedure. Otherwise, for the numbers from 20 to 99,
we do this.
digit_names3 = [
"twenty", "thirty" , ....... "eighty", "ninety"
]
We take first digit, get the name from the digit_names3.
Digit 1 = digit_names3 [number//10-2]
add hypen and add digit names [number%10]if the second digit is not zero. Else, add nothing.
## Integers under 1000
Digit 1 = digit_names [number//100]
add "hundred"
R1 = number%100
If R1 is not zero, we add "and", then we call Integers under 100 procedure on R1.
## Integers under 1000000
Q1 = number//1000
We call Intergers under 1000 procedure on Q1
We add "thousand"
R1 = number%1000
If R1 is not zero, we call Integers under 1000 procedure on R1
## Negative integers down to -1 million
We won't deal with negative integers in this problem set,
but how would you deal with a negative integer, using the
functions above?
Number multiped by negative 1, we treat it as a new number.
We call the same function as before (now it's positive number)
We write "negative" in the begining of the answer.