Personally, i would use a 3-D printer, but you can use any material that you have, and have an example next to you for reference if you aren't going to 3-D print.
Answer:
Check the explanation
Explanation:
Here in this game of Tic-Tac-Toe, it is using the TPGE engine which is a Tiny Python Game Engine. Let's talk about its functions like:-
def image_type(img): In this function, it is simply taking image as a parameter and returning its object type like DISC if the image is in graphical form, TEXT if it is a string, and LINE if it is other than the mentioned object.
def convert_image(img): In this function, it is simply taking image as a parameter and returning an equivalent graphical object as understood by John Zelle's. Mainly comparing for three things in this function and those are: if image equals to DISC then it is calling convert_circle(function), if image equals to LINE then it is calling convert_line(function), and if image equals to TEXT then it is calling convert_text(function),
def convert_circle(x): This function takes a list( a group of values) and makes a circle at the center of the window and the circle's radius is coming from the list.
convert_text, convert_line, convert_circle are only creating text, line, and circle and then returning it.
def graphical_elements(images): This function is taking image as a parameter and then extracting shape and color from the image and then calling convert_image(shape) and convert_type(shape) and it gives us graphic and kind respectively. Now it is checking whether kind equals to DISC If yes then filling color on the window else, setting the outline of the window
def run(): Here it is finally running the game with required parameters, the whole game is continously running under the while loop.
That's all
Answer:
Check the explanation
Explanation:
10
5
3
21
2
-6
the output will be somthing like this:
2 21
You can then suppose that the list of integers will contain at least 2 values.
Answer:
int x;
indata.open("lottowins");
indata >> x;
cout << x << endl;
indata >> x;
cout << x << endl;
indata >> x;
cout << x << endl;
indata.close();
