用c语言解决下列问题编写一个程序，输入年份和月份，判断该年是否是闰年，并根据给出的月份判断是什么季节和该月有多少天？（闰年的条件是年份能被4整除但不能被100整除，或者能被400整除；规定3~5月为春季，6~8月为夏季，9~11月为秋季，1、2和12月为冬季）。 **输入格式要求："%d,%d" 提示信息："Please enter year,month:" **输出格式要求："%d is leap year\n" "%d is not leap year\n" "The season is spring\n" "The season is summer\n" "The season is autumn\n" "The season is winter\n" "The number of days of this month is %d\n" 程序运行示例如下： 实例1： Please enter year,month:2012,11 2012 is leap year The season is autumn The number of days of this month is 30 实例2： Please enter year,month:2013,12 2013 is not leap year The season is winter The number of days of this month is 31

好的，下面是代码实现：

#include <stdio.h>

int main()
{
    int year, month, days;
    printf("Please enter year,month:");
    scanf("%d,%d", &year, &month);

    // 判断是否为闰年
    if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
    {
        printf("%d is leap year\n", year);
    }
    else
    {
        printf("%d is not leap year\n", year);
    }

    // 判断季节
    if (month >= 3 && month <= 5)
    {
        printf("The season is spring\n");
    }
    else if (month >= 6 && month <= 8)
    {
        printf("The season is summer\n");
    }
    else if (month >= 9 && month <= 11)
    {
        printf("The season is autumn\n");
    }
    else
    {
        printf("The season is winter\n");
    }

    // 判断该月有多少天
    switch (month)
    {
    case 2:
        if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
        {
            days = 29;
        }
        else
        {
            days = 28;
        }
        break;
    case 4:
    case 6:
    case 9:
    case 11:
        days = 30;
        break;
    default:
        days = 31;
        break;
    }
    printf("The number of days of this month is %d\n", days);

    return 0;
}

这样，就可以实现输入年份和月份，判断该年是否为闰年，判断季节，以及该月有多少天的功能了。