#include <stdio.h>
int main()
{
/* 定义需要计算的日期 */
int year = 2008;
int month = 8;
int day = 8;
/*
* 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~
*/
int sum = 0;
switch(month)
{
case 1:
sum = 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("输入错误,一年只有12个月");
}
if(month > 2 && year%4 == 0 && year%100 != 0) //判断是否是闰年
{
sum = sum + 1; //闰年的2月份比平年多一天
}
printf("%d年%d月%d日这一天，是该年的第%d天。", year,month,day,sum);
return 0;
}
#include <stdio.h>
int main()
{
int year = 2008, sum = 0;
int month = 8;
int day = 8;
switch(month-1)
{
case 12:sum+=31;
case 11:sum+=30;
case 10:sum+=31;
case 9:sum+=30;
case 8:sum+=31;
case 7:sum+=31;
case 6:sum+=30;
case 5:sum+=31;
case 4:sum+=30;
case 3:sum+=31;
case 2:sum+=28;
case 1:sum+=31;
default:
if(year%4 == 0 && year%100 != 0 && month > 2)
{
sum+=(1+day);
}
printf("%d年%d月%d日是该年的第%d天！", year, month, day, sum);
break;
}
return 0;
}
#include <stdio.h>
int main()
{
int month,day;
int day1=8;
for(month=1,day=day1;month<8;month++)
{
if(month%2==1)
{
day+=31;
}
else
{
if(month==2)
{
day+=29;
continue;
}
day+=30;
}
}
printf("2008年%d月%d日是该年的第%d天",month,day1,day);