华为OD机试

OD机试

前两题忘了，很简单，但是ac都只有90
第三题ac 25，能想到的用例都能通过，可能是我没有读懂题
有没有大佬来指正一下

输入ABV
连续字符串满足 在A和B中的位置相同并且连续
连续字符串的ASCLL绝对值之和|A[i]-B[i]|之和小于等于V
输出最长的连续字符串数量
（略其它无关的限制条件）
# 示例输入
# xxabcdef
# aacdefgh
# 5
# 输出
# 2
# 说明：ab->cd，bc->de,cd->ef,...最长的为2

A = input()
B = input()
V = int(input())
assert 1 <= len(A) == len(B) <= 1000 and A.islower() and B.islower()
pd_1 = [0 for _ in A]
pd_2 = [0 for _ in A]
pd_3 = [abs(ord(A[i]) - ord(B[i])) for i in range(len(A))]
for i, v in enumerate(A[:-1]):
    if chr(ord(v) + 1) == A[i + 1] and chr(ord(B[i]) + 1) == B[i + 1]:
        if pd_1[i] == 0:
            pd_1[i] = 1
        pd_1[i + 1] = pd_1[i] + 1
for i, v in enumerate(pd_1):
    if v != 0:
        if i == 0:
            pd_2[i] = pd_3[i]
        elif v == pd_1[i - 1] + 1:
            pd_2[i] = pd_3[i] + pd_2[i - 1]
        elif v != 0:
            pd_2[i] = pd_3[i]

max_len = 0
for i, v in enumerate(pd_1):
    if v == 0:
        continue
    if pd_2[i] <= V:
        max_len = max(max_len, v)
if max_len == 0:
    min_abs = min(pd_3)
    max_len = 1 if min_abs <= V else 0

print(max_len)