查找斐波那契一致子数组的总和

Question

We have an input integer let's say 13. We can find consistent subarray of fibonacci numbers that sums to 10 - [2,3,5]. I need to find next number that is not a sum of consistent subarray. In this case this number will be 14. I have this code, but the catch is, it can be optimized to not iterate through all of the N's from starting Left Pointer = 1 and Right Pointer = 1 but somehow "import" from previous N and i have no clue how to do it so maybe someone smarter might help.

def fib(n):
    if n == 1: return 1
    if n == 2: return 1
    return fib(n-1) + fib(n-2)


def fibosubarr(n):
    L_pointer = 1
    R_pointer = 2
    sumfibs = 1
    while sumfibs != n:

        if sumfibs > n and L_pointer < R_pointer:
            sumfibs -= fib(L_pointer)
            L_pointer += 1

        elif sumfibs < n and L_poiner < R_pointer:
            sumfibs += fib(R_pointer)
            R_pointer += 1

        else: return False
    return True
    
n = int(input())

while fibosubarr(n):
    n += 1
print(n)

Answer 1

Here's a technique called "memoizing". The fib function here keeps track of the current list and only extends it as necessary. Once it has generated a number, it doesn't need to do it again.

_fib = [1,1]
def fib(n):
    while len(_fib) <= n:
        _fib.append( _fib[-2]+_fib[-1] )
    return _fib[n]

With your scheme, 200000 caused a noticeable delay. With this scheme, even 2 billion runs instantaneously.

Answer 2

To get the next subarray sum, you only need one call of the function -- provided you keep track of the least sum value that was exceeding n .

I would also use a generator for the Fibonacci numbers:

def genfib():
    a = 1
    b = 1
    while True:
        yield b
        a, b = b, a + b

def fibosubarr(n):
    left = genfib()
    right = genfib()

    sumfibs = next(right)
    closest = float("inf")
    while sumfibs:
        if sumfibs > n:
            closest = min(sumfibs, closest)
            sumfibs -= next(left)
        elif sumfibs < n:
            sumfibs += next(right)
        else:
            return n
    return closest

Now you can do as you did -- produce the next valid sum that is at least the input value:

n = int(input())
print(fibosubarr(n))

You could also loop to go from one such sum to the next:

n = 0
for _ in range(10):  # first 10 such sums
    n = fibosubarr(n+1)
    print(n)