import re
def count_num(start, end):
    pattern = re.compile(r'.*2.*0.*2.*3.*')  # 正则表达式模式，匹配包含子串"2023"的任意字符
    count = 0
    for num in range(start, end + 1):
        if not pattern.search(str(num)):  # 如果数字不匹配正则表达式模式
            count += 1
    return count
start = 12345678
end = 98765432
result = count_num(start, end)
print(f"在 {start} 至 {end} 中，有 {result} 个数中完全不包含 2023。")