The Solution

def fib(n):

if n == 1:

return 0

if n == 2:

return 1

numbers = [0, 1]

for i in range(n-2):

new = numbers[-2] + numbers[-1]

numbers.append(new)

return numbers[-1]

for i in range(1, 11):

print(fib(i), end=' ')

This code initializes a list with the first two Fibonacci numbers, [0, 1]. We generate new Fibonacci numbers by adding the last two numbers in the list and appending the result. This process is repeated n-2 times, with the final number in the list representing the nth Fibonacci number.

For a more space-efficient approach, consider the following implementation:

def fib(n):

if n == 1:

return 0

if n == 2:

return 1

a = 0

b = 1

for i in range(n-2):

c = a + b

a = b

b = c

return c

for i in range(1, 11):

print(fib(i), end=' ')

In this version, we use three variables: a, b, and c. We compute the new Fibonacci number with c = a + b and shift a and b forward. This reduces the space complexity while still allowing us to calculate the Fibonacci numbers efficiently.

Conclusion

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