题解 | #字符串通配符#

str1 = input().lower()
str2 = input().lower()


# 初始化动态规划矩阵
dp = [[False for _ in range(len(str2)+1)] for _ in range(len(str1)+1)]
dp[0][0] = True
for i in range(1, len(str1)+1):
    if dp[i-1][0] and str1[i-1] == '*':
        dp[i][0] = True
    else:
        dp[i][0] = False


# 动态规划求解
for i in range(1, len(str1)+1):
    for j in range(1, len(str2)+1):
        if str1[i-1] == '?':
            if (dp[i - 1][j - 1] or dp[i][j-1]) and (str2[j - 1].isalpha() or str2[j - 1].isdigit()):
                dp[i][j] = True
            else:
                dp[i][j] = False
        elif str1[i-1] == '*':
            if dp[i-1][j] or (dp[i][j-1] and (str2[j - 1].isalpha() or str2[j - 1].isdigit())):
                dp[i][j] = True
            else:
                dp[i][j] = False
        else:
            if dp[i - 1][j - 1] and str1[i - 1] == str2[j - 1]:
                dp[i][j] = True
            else:
                dp[i][j] = False


if dp[len(str1)][len(str2)]:
    print('true')
else:
    print('false')