how to print star hypen pattern in python-CodePudding

How to print the pattern like this:

(need to print n-1 lines)

input=3

----@
--@-@-@

input=6

----------@
--------@-@-@
------@---@---@
----@-----@-----@
--@-------@-------@

My code:

row = int(input())
for i in range(1, row):
    for j in range(1,row-i 1):
        print("-", end="")
    for j in range(1, 2*i):
        if j==1 or j==2*i-1:
            print("@", end="")
        else:
            print("-", end="")
    print()

MY OUTPUT: input=5

----@
---@-@
--@---@
-@-----@

Please explain how to do??

CodePudding user response：

There are a few things missing and to be improved in your code:

There's no need to make a loop to print the same character again and again: on python you can use the product to repeat the character an x number of times. For example: "-" * 3 == "---"
They way you calculate the hyphens in the middle is fine, but you need to do it twice and add an "@" in between.
You can build the strings part by part first and then print the whole line, avoiding having to print an empty line in the end of the loop.
Personally, since the first line is going to have one "@" and not three, I prefer to calculate it and print it separately.

With these improvements, a solution to your problem could be:

row = int(input())
print("-" * (row - 1) * 2   "@")
for i in range(row - 2, 0, -1):
    left_hyphens = "-" * i * 2
    mid_hyphens = "-" * (1   2 * (row - 2 - i))
    print(left_hyphens   "@"   mid_hyphens   "@"   mid_hyphens   "@")

CodePudding user response：

row = int(input())
for i in range(1, row):
    for j in range(1,2*(row-i) 1):
        print("-", end="")
    for j in range(1, 4*i):
        if j==1 or j==2*i-1 or j==4*i-3:
            print("@", end="")
        elif j<=4*i-3:
            print("-", end="")
    print()