.编程实现NextDate问题，并给出相应的测试用例 NextDate是一个有三个变量(月份、日期和年)的函数。函数返回输入日期后面的那个日期。变量月份、日期和年都是整数，且满足以下条件: C1:1<

编程实现`NextDate`问题，你可以创建一个简单的Python函数来完成这个任务，假设我们使用普通的业务规则，即一个月份最多有30天（闰年的二月有29天），并且考虑平闰年的转换。这里是一个基本的实现：

def is_leap_year(year):
    return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

def days_in_month(month, year):
    if month in [1, 3, 5, 7, 8, 10, 12]:  # 除2月外的非闰月
        return 30
    elif month == 2:
        if is_leap_year(year):
            return 29
        else:
            return 28
    else:  # 其他闰月
        return 31

def next_date(month, date, year):
    next_month = (month + 1) % 13  # 转换到下个月，因为每个月加上1可能会超过12
    days_left_in_current_month = days_in_month(month, year)
    
    # 如果下个月的第一天不是当前日期，则需要加一天
    if date > 1:
        next_day = (date - 1) + days_left_in_current_month
    else:
        next_day = days_left_in_current_month
    
    return next_month, next_day, year

# 测试用例
test_cases = [
    (1, 31, 2022),  # February 28th, 2022 -> March 1st, 2022
    (3, 30, 2022),  # March 30th, 2022 -> April 1st, 2022
    (6, 30, 2022),  # June 30th, 2022 -> July 1st, 2022
    (9, 30, 2022),  # September 30th, 2022 -> October 1st, 2022
    (12, 31, 2022),  # December 31st, 2022 -> January 1st, 2023 (leap year)
]

for case in test_cases:
    month, date, year = case
    result = next_date(month, date, year)
    print(f"Input: {month}/{date}/{year}, Next Date: {result[0]}/{result[1]}/{result[2]}")