#include <stdio.h>

int main() 

{ 

    /* 定义需要计算的日期 */

    int year = 2008;

    int month = 8;

    int day = 8;

    int sum,flag;

    /*

     * 请使用switch语句，if...else语句完成本题

     * 如有想看小编思路的，可以点击左侧任务中的“不会了怎么办”

     * 小编还是希望大家独立完成哦~

     */

    switch(month)

    {

        case 1:sum=0+day;break;

    case 2:sum=31+day;break;

    case 3:sum=31+28+day;break;

    case 4:sum=31+28+31+day;break;

    case 5:sum=31+28+31+30+day;break;

    case 6:sum=31+28+31+30+31+day;break;

    case 7:sum=31+28+31+30+31+30+day;break;

    case 8:sum=31+28+31+30+31+30+31+day;break;

    case 9:sum=31+28+31+30+31+30+31+31+day;break;

    case 10:sum=31+28+31+30+31+30+31+31+30+day;break;

    case 11:sum=31+28+31+30+31+30+31+31+30+31+day;break;

    case 12:sum=31+28+31+30+31+30+31+31+30+31+30+day;break;

    default :printf("一年中只有12个月");break;

    }

    if((year%400==0)||((year%4==0)&&(year%100!=0)))&&(month>2)

    {

    sum=sum+1;

    printf("today is the %d day in the year",sum);

}

    else

    printf("today is the %d day in the year",sum);

return 0;

}