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creating chessboard pattern with tables in python

I need to make a chessboard pattern filled with 0 and 1 but it it doesn't have to be square table
I need to get rows and columns from user
just an example:
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
I have the solution but I couldn't understand the last row of code (table[i][j] = int(not table[i][j-1]))
can we solve it using another method?
m = int(input("insert number of rows: "))
n = int(input("insert number of colomns: "))
table = list()
for i in range(0,m):
if i%2 == 0 :
table[i][0] = 1
else:
table[i][0] = 0
for j in range(1,n):
table[i][j] = int(not table[i][j-1])
print("print the table with chessboard pattern")
for i in range(0,m):
for j in range(0,n):
print(table[i][j],end='')
print()

You'll have a simpler time with a nested list comprehension that uses the x and y coordinates of the cell being generated and the modulo operator % to alternate between two values:
n_rows = n_cols = 5
table = [
[((x + y + 1) % 2) for x in range(n_cols)]
for y in range(n_rows)
]
for row in table:
print(*row)
prints out
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
The list comprehension can be written out as nested for loops, too.
table = []
for y in range(n_rows):
row = []
for x in range(n_cols):
row.append((x + y + 1) % 2)
table.append(row)

You will have an error on line 6 because you can't assign a value to an index that doesn't exist..
You can do this instead:
m = int(input("insert number of rows: "))
n = int(input("insert number of colomns: "))
table = []
for i in range(m):
if i%2==0:
table.append([0+j%2 for j in range(n)])
else:
table.append([0+j%2 for j in range(1, n+1)])
You could also print it like this:
print("".join([str(row).replace('[', '').replace(']', '\n').replace(', ', ' ') for row in table]))

How to receive a num at each step and continue until zero is entered; then this program should print each entered num as its own num

Here I have this code:
N = int(input())
Tmp=n
While tmp>0 :
Print(n)
Tmp-=1
But for ex: when I have:
3
2
1
0
As entered nums, it just prints:
3
3
3
But I need to print:
3
3
3
2
2
1
Here I have this code:
N = int(input())
Tmp=n
While tmp>0 :
Print(n)
Tmp-=1
But there is a problem!bc it just prints
3
3
3
Instead of:
3
3
3
2
2
1

you need to print(tmp) instead of print(n)
this will print 3 2 1.
To get 3 3 3 2 2 1 you need to change your code more:
n = int(input())
tmp=n
while tmp > 0:
for _ in range(tmp)
print(tmp)
tmp -= 1

You need to add if statement when tmp equals 0 then subtract n by 1 and assign tmp back to n:
n = int(input('input : '))
tmp = n
while tmp > 0:
print(n)
tmp -= 1
if tmp == 0:
n -= 1
tmp = n
Output:
# input : 3
# 3
# 3
# 3
# 2
# 2
# 1

How to get the count of lines in a output of a python script

This is my code for getting a 3 digit combination.
a = input("Enter first number: ")
b = input("Enter second number: ")
c = input("Enter third number: ")
d = []
d.append(a)
d.append(b)
d.append(c)
for i in range(0, 4):
for j in range(0, 4):
for k in range(0, 4):
if(i !=j & j!=k & k!=i):
for count, i in enumerate(range(4),1):
print(i,j,k)
Input:
1,2,3
and
Output:
0 1 2
1 1 2
2 1 2
3 1 2
0 2 0
1 2 0
2 2 0
3 2 0 & more....
My question is how can I get the count of how many combinations I have got?
Thank you so much for your attention and participation.

Outside of your loop, create an integer variable and set it equal to 0. Where you print out the combinations (within the loop), add 1 to this integer variable. Finally at the end of your program outside the loop, print the value of the counter variable.
For example:
counter = 0
for i in range(0, 4):
for j in range(0, 4):
for k in range(0, 4):
if (i != j & j != k & k != i):
for count, i in enumerate(range(4), 1):
print(i, j, k)
counter += 1
print(counter)

Find the sum of all the multiples of 3 or 5 below 1000. (idk what is wrong with my code)

I'm a beginner and I tried this code to list the sum of all the multiples of 3 or 5 below 100, but it gives a wrong answer and I don't know why.
result = 0
result2 = 0
som = 0
sum2 = 0
below_1000 = True
while below_1000:
result = result+3
if result < 1000:
som += result
else:
below_1000 = False
below_1000 = True
while below_1000:
result2 = result2+5
if result2 < 1000:
sum2 += result2
else:
below_1000 = False
final_result = som+sum2
print(final_result)

Since you first loop over multiples of 3, then again over multiples of 5, you are double-counting a lot of values, specifically values that are multiples of both 3 and 5 (for example 15 or 60).
To write this manually, you can use a for loop over range
total = 0
for i in range(1000):
if i % 3 == 0 or i % 5 == 0:
total += i
>>> total
233168
A more concise way to do this same thing is using a generator expression within the sum function
>>> sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)
233168

How do I reduce the number of loops or complexity

So what I am trying to do is to find count of alternating numbers such that it alternates with -ve and positive sign
for eg: 1 -2 3 -4 would get me 4 3 2 1 as from 1 to -4 including the two numbers there are 4 numbers.
Simillarly for 1 1 -3 2 would get me 1 3 2 1
Now I have the code but I cannot optimise it and it returns me a time limit exceeded error even though it works for moderate input stream.
j=0
count=0
length=(raw_input())
st=map(int,raw_input().split())
while j+1 < len(st):
k=j+1
count=0
temp=j
while k<len(st) and ((st[k]<0 and st[j]>0) or (st[k]>0 and st[j]<0)):
count+=1
k+=1
j+=1
print count+1,
j=temp+1
print 1

Try using for loops instead of while loops as that avoids you some variable assignments:
st = map(int, raw_input().split())
length = len(st)-1
for i in range(length):
count = 1
for j in range(i, length):
if (st[j]<0 and st[j+1]>0) or (st[j+1]<0 and st[j]>0):
count += 1
else:
break
print(count)
print(1)
This will give:
<< 1 -2 3 4
>> 4
>> 3
>> 2
>> 1
<< 1 1 -3 2
>> 1
>> 3
>> 2
>> 1
It may also be a bit faster if you extract the numbers from the list once instead of twice:
st = map(int, raw_input().split())
length = len(st)-1
for i in range(length):
count = 1
for j in range(i, length):
first, second = st[j:j+2]
if (first<0 and second>0) or (first>0 and second<0):
count += 1
else:
break
print(count)
print(1)
The last thing I would try is checking that they sigs are different with a single comparisson but I do not really expect this to be faster:
st = map(int, raw_input().split())
length = len(st)-1
for i in range(length):
count = 1
for j in range(i, length):
product = st[j] * st[j+1]
if product != abs(product):
count += 1
else:
break
print(count)
print(1)