D: Doubleplusgood

https://open.kattis.com/problems/doubleplusgood

Since there are at most 18 digits in the input, there can be at most 8 plus signs, so the problem is brute-forceable in at most $2^8=256$ evaluations of the input expression.

To brute force search over all binary strings of length $n$, we increment a counter $i$ from $0$ to $2^n-1$; the $j$th digit in the binary string is then $0$ if $(2^j) \& i = 0$ and $1$ otherwise, where $\&$ is the bitwise AND function. For a given binary string, we then evaluate the input by using the binary string to decide which symbols to interpret as $\boxplus$. Since $\boxplus$ has higher precedence than $+$, we iterate from left to right, keeping track of the sum of previous concatenated terms and the current number; if the next operation is $\boxplus$, we concatenate the next number to the current number, and if the next operation is $+$, then we add the current number to the running sum and reset the current number to $0$. We use a set to keep track of the distinct values that we obtain trying all binary strings. In code:

ints = input().split('+')
lens = list(map(len, ints))
ints = list(map(int, ints))
n = len(ints) - 1
possibilities = set()
for i in range(1 << n):
    curr_num = ints[0]
    curr_sum = 0
    for j in range(n):
        if (1 << j) & i:
            curr_num = curr_num * pow(10, lens[j+1]) + ints[j+1]
        else:
            curr_sum += curr_num
            curr_num = ints[j+1]
    curr_sum += curr_num
    possibilities.add(curr_sum)

print(len(possibilities))