我的代码如下：
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))
以下案例使用此算法失败：
10 10我的输出：4，正确的输出：5
10 200我的输出：5，正确的输出：2
1234 12345我的输出：2，正确的输出：8
（可能还有更多）
我想知道问题出在哪里，如何解决？有没有使用相同基础的更好方法？
循环数存在问题：您在应该存在x的地方执行x + 1循环。而且我不明白您为什么不以sum = 0开头。
然后，您可以利用周期以恒定时间计算总和，而无需任何循环。
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)))
还请注意，在最后一步中，由于计算mod 10的差异可能为负，因此应该为：
def fibtotal(m, n):
return (fibtotal3(n) - fibtotal3(m - 1)) % 10
对于经过的读者：fibtotal2和fibtotal3起作用是因为fib(n) % 10是周期为60的周期，并且周期的元素之和是10的倍数。请参见Math.SE上的Fibonacci's final digits cycle every 60 numbers。