#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;
switch(month)
{
case 1:sum=0;
break;
case 2:sum=31;
break;
case 3:sum=31+28;
break;
case 4:sum=31+28+31;
break;
case 5:sum=31+28+31+30;
break;
case 6:sum=31+28+31+30+31;
break;
case 7:sum=31+28+31+30+31+30;
break;
case 8:sum=31+28+31+30+31+30+31;
break;
case 9:sum=31+28+31+30+31+30+31+31;
break;
case 10:sum=31+28+31+30+31+30+31+31+30;
break;
case 11:sum=31+28+31+30+31+30+31+31+30+31;
break;
case 12:sum=31+28+31+30+31+30+31+31+30+31+30;
break;
default:
printf("一年只有十二个月哟\n");break;
}
sum+=day;
if(year/4==0&&year/100!=0||year/400==0)
{
if(month>=2)
{
sum++;
day+=sum;
}
else
day+=sum;
printf("%d年的第%d天\n",year,day);
}
/*
* 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~
*/
return 0;
}
#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;
switch(month)
{
case 1:sum=0;
break;
case 2:sum=31;
break;
case 3:sum=31+28;
break;
case 4:sum=31+28+31;
break;
case 5:sum=31+28+31+30;
break;
case 6:sum=31+28+31+30+31;
break;
case 7:sum=31+28+31+30+31+30;
break;
case 8:sum=31+28+31+30+31+30+31;
break;
case 9:sum=31+28+31+30+31+30+31+31;
break;
case 10:sum=31+28+31+30+31+30+31+31+30;
break;
case 11:sum=31+28+31+30+31+30+31+31+30+31;
break;
case 12:sum=31+28+31+30+31+30+31+31+30+31+30;
break;
default:
printf("一年只有十二个月哟\n");break;
}
if((year%4==0&&year%100!=0)||year%400==0)
{
if(month>=2)
{
sum+=1;
day+=sum;
printf("2008年8月8日是该年的第%d天",day);
}
}
else
{
day+=sum;
printf("2008年8月8日是该年的第%d天",day);
}
/*
* 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~
*/
return 0;
}