[编程入门]结构体之时间设计-题解（C语言代码）

解题思路:思绪有点乱，大概就这样了

注意事项:

参考代码:

#include

int main()

{

    int year,month,day;

    scanf("%d %d %d",&year,&month,&day);

    if(year%100==0)//此处循环是为了判断能否被100整除

    {

        if(year%4==0)//如果既能被100整除又能被4整除就是闰年

        {

            switch(month)//此处用swich和case方便列举

            {

                case 1:printf("%d",day);break;//每个case语句结束必须跟上break

                case 2:printf("%d",31+day);break;//闰年二月29天

                case 3:printf("%d",60+day);break;

                case 4:printf("%d",91+day);break;

                case 5:printf("d",121+day);break;

                case 6:printf("%d",152+day);break;

                case 7:printf("%d",182+day);break;

                case 8:printf("%d",213+day);break;

                case 9:printf("%d",244+day);break;

                case 10:printf("%d",274+day);break;

                case 11:printf("%d",305+day);break;

                case 12:printf("%d",335+day);break;

            }

        }

        if(year%4!=0)//如果能被100整除而不能被4整除就不是闰年，就得按平年计算日子数

        {

              switch(month)

            {

                case 1:printf("%d",day);break;

                case 2:printf("%d",31+day);break;//平年二月28天

                case 3:printf("%d",59+day);break;

                case 4:printf("%d",90+day);break;

                case 5:printf("d",120+day);break;

                case 6:printf("%d",151+day);break;

                case 7:printf("%d",181+day);break;

                case 8:printf("%d",212+day);break;

                case 9:printf("%d",243+day);break;

                case 10:printf("%d",273+day);break;

                case 11:printf("%d",304+day);break;

                case 12:printf("%d",334+day);break;

            }

        }

    }

    if(year%100!=0)//这个函数是判断能否被100整除，不能被100整除则进入该函数

    {

    if(year%4!=0)//判断不能被100整除是否能被4整除，如果不能被4整除就按照平年计算天数

    {

          switch(month)

            {

                case 1:printf("%d",day);break;

                case 2:printf("%d",31+day);break;

                case 3:printf("%d",59+day);break;

                case 4:printf("%d",90+day);break;

                case 5:printf("d",120+day);break;

                case 6:printf("%d",151+day);break;

                case 7:printf("%d",181+day);break;

                case 8:printf("%d",212+day);break;

                case 9:printf("%d",243+day);break;

                case 10:printf("%d",273+day);break;

                case 11:printf("%d",304+day);break;

                case 12:printf("%d",334+day);break;

            }

    }

    if(year%4==0)//如果能被4整除就按照闰年来计数

    {

        switch(month)

            {

                case 1:printf("%d",day);break;

                case 2:printf("%d",31+day);break;

                case 3:printf("%d",60+day);break;

                case 4:printf("%d",91+day);break;

                case 5:printf("d",121+day);break;

                case 6:printf("%d",152+day);break;

                case 7:printf("%d",182+day);break;

                case 8:printf("%d",213+day);break;

                case 9:printf("%d",244+day);break;

                case 10:printf("%d",274+day);break;

                case 11:printf("%d",305+day);break;

                case 12:printf("%d",335+day);break;

            }

    }

    }

    return 0;//函数结束

}