Python: Calculate the last digit of a large Fibonacci Number with less time

# Uses python3

# Compute the Last Digit of a Large Fibonacci Number
def Fib_Last_Digit(n):

    if n == 0 : return 0
    elif n == 1: return 1
    else:

        a,b = 0,1
        for i in range(1,n):
            c = a + b;
            a = b;
            b = c;
        # Calculate the last digit of the final number 
        lastdigit = int(repr(c)[-1]);
        print(lastdigit);

n = int(input(""));
Fib_Last_Digit(n);

This code works very well. However, I want to revise the algorithm to save more time and memory. By the way, the input and output should be kept the same with the previous version.

Only keeping the last digit during calculation saves a lot of time:

def fib_last_digit(n):
    if n < 2: return n
    else:
        a, b = 0, 1
        for i in range(1,n):
            a, b = b, (a+b) % 10
        print(b)

n = int(input())
fib_last_digit(n)

Handling numbers that fit in fewer bytes saves time.

When you're working with huge numbers, you can save a lot of time using the answer described here, slightly modified to only keep track of the last digit:

def fib_last_digit(n):
    v1, v2, v3 = 1, 1, 0    # initialise a matrix [[1,1],[1,0]]
    for rec in bin(n)[3:]:  # perform fast exponentiation of the matrix (quickly raise it to the nth power)
        calc = (v2*v2) % 10
        v1, v2, v3 = (v1*v1+calc) % 10, ((v1+v3)*v2) % 10, (calc+v3*v3) % 10
        if rec == '1': v1, v2, v3 = (v1+v2) % 10, v1, v2
    return v2

And finally, based on the concept described in Eugene Yarmash's answer (the last digits repeat every 60 steps) we can make an even faster solution (O(1)):

def fib_last_digit(n):
    return (
        [1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, 1, 7, 8, 5, 3, 8, 1, 9, 0, 9, 9, 8, 7, 5, 2, 7, 9, 6, 5, 1, 6, 7, 3, 0, 3, 3, 6, 9, 5, 4, 9, 3, 2, 5, 7, 2, 9, 1, 0]
        [n % 60 - 1]
    )