Finding last digit of sum from m to n Fibonacci numbers. (0 ≤ 𝑚 ≤ 𝑛 ≤ 10^14)

My code is as follow :

m, n = map(int, input().split())

# write function "fibtotal" which takes input x and gives accurate fib(x+2)%10  (as sum till fib(x) == fib(x+2) - 1)
# using above function get fibtotal(m-1) and fibtotal(n)
# subtract fibtotal(m-1) from fibtotal(n) and do mod 10 gives last digit of sum from m to n
# take care of handling large input sizes, 0 ≤ 𝑚 ≤ 𝑛 ≤ 10^14

def fibtotal(x):

  sum = 1 # if both initial conditions fail then loop starts from 2

  x= x % 60 # pisano period of 10 is 60 and to get last digit we need to divide by 10

  if x == 0:
    sum = 1 # fib(2)
    return sum 

  if x == 1:
    sum = 2 # fib(3)
    return sum

  a, b = 0, 1

  for i in range(2, x+3): # to find sum till fib(x+2)

    c = (a+b)%10
    sum += c
    a, b = b%10, c%10

  return sum%10

# no need to subtract 1 from both as they cancel out
print(fibtotal(n)-fibtotal(m-1))

Following Cases fail using this algorithm:

10 10 My output: 4, correct output: 5

10 200 My output: 5, correct output: 2

1234 12345 My output: 2, correct output: 8

(and possibly many more)

I want to know where is the problem and how can I fix it? Is there any better approach using same fundamentals?

There is a problem in the number of loop: you do x+1 loops where there should be x. And I don't understand why you don't start with sum = 0.

Then, you can make use of the period to compute the sum in constant time, without any loop. The aux list was computed using fibtotal1.

def fib(n):
    a, b = 0, 1
    for i in range(n):
        a, b = b, a + b
    return a

def fibtotal1(n):
    return sum(fib(k) % 10 for k in range(n + 1)) % 10

def fibtotal2(n):
    s, a, b = 0, 0, 1
    for i in range(n % 60):
        a, b = b, a + b
        s += a
    return s % 10

aux = [0, 1, 2, 4, 7, 2, 0, 3, 4, 8, 3, 2, 6, 9, 6, 6, 3, 0, 4, 5,
       0, 6, 7, 4, 2, 7, 0, 8, 9, 8, 8, 7, 6, 4, 1, 6, 8, 5, 4, 0,
       5, 6, 2, 9, 2, 2, 5, 8, 4, 3, 8, 2, 1, 4, 6, 1, 8, 0, 9, 0]

def fibtotal3(n):
    return aux[n % 60]

print(all(fibtotal1(n) == fibtotal2(n) == fibtotal3(n) for n in range(1000)))

Note also that in your last step, due to computing mod 10 the difference may be negative, so it should be:

def fibtotal(m, n):
    return (fibtotal3(n) - fibtotal3(m - 1)) % 10

For the reader passing by: fibtotal2 and fibtotal3 work because fib(n) % 10 is periodic with period 60, and the sum of the elements of the period is a multiple of 10. See Fibonacci's final digits cycle every 60 numbers on Math.SE.