为什么会运行失败
錯U
2020-08-12 20:33
#include <stdio.h>
int main()
{
/* 定义需要计算的日期 */
int year = 2008;
int month = 8;
int day = 8;
int i,sum;
/*
* 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;break;
case 2:sum=31;break;
case 3:sum=31+29;break;
case 4:sum=31+29+31;break;
case 5:sum=31+29+31+30;break;
case 6:sum=31+29+31+30+31;break;
case 7:sum=31+29+31+30+31+30;break;
case 8:sum=31+29+31+30+31+30+31;break;
case 9:sum=31+29+31+30+31+30+31+31;break;
case 10:sum=31+29+31+30+31+30+31+31+30;break;
case 11:sum=31+29+31+30+31+30+31+31+30+31;break;
case 12:sum=31+29+31+30+31+30+31+31+30+31+30;break;
default:printf("一年12个月")；break;
}
sum=sum+day;
if(year%400==0)
i=1;
else
i=0;
if(i==0&&month>2)
sum--;
printf("%d年%月%日是该年的第%d天"，year,month,day,sum)；
return 0;
}
2回答
-
- 周纳尔多
- 2020-08-16 21:36:05
#include <stdio.h>int main() { int year = 2008; int month = 8; int day = 8; int i,j; if(year%100==0&&year%400==0||year%4==0&&year%100!=0)i=1; else i=0; switch(month) { case 1:j=day;break; case 2:j=31+ day;break; case 3:j=31+ 28+ day;break; case 4:j=31+ 28+ 31+ day;break; case 5:j=31+ 28+ 31+ 30+ day;break; case 6:j=31+ 28+ 31+ 30+ 31+ day;break; case 7:j=31+ 28+ 31+ 30+ 31+ 30+ day;break; case 8:j=31+ 28+ 31+ 30+ 31+ 30+ 31+day;break; case 9:j=31+ 28+ 31+ 30+ 31+ 30+ 31+ 31+ day;break; case 10:j=31+ 28+ 31+ 30+ 31+ 30+ 31+ 31+ 30+day;break; case 11:j=31+ 28+ 31+ 30+ 31+ 30+ 31+ 31+ 30+ 31+day;break; case 12:j=31+ 28+ 31+ 30+ 31+ 30+ 31+ 31+ 30+ 31+ 30+ day;break; } if(i==1)j+=1; printf("%d年%d月%d日是该年的第%d天",year,month,day,j); return 0;} -
- 慕圣5070646
- 2020-08-13 11:17:49
#include <stdio.h>
int main()
{
int year = 2008;
int month = 8;
int day = 8;
int i,j;
if(year%100==0&&year%400==0||year%4==0&&year%100!=0)i=1;
else i=0;
switch(month)
{
case 1:j=day;break;
case 2:j=31+ day;break;
case 3:j=31+ 28+ day;break;
case 4:j=31+ 28+ 31+ day;break;
case 5:j=31+ 28+ 31+ 30+ day;break;
case 6:j=31+ 28+ 31+ 30+ 31+ day;break;
case 7:j=31+ 28+ 31+ 30+ 31+ 30+ day;break;
case 8:j=31+ 28+ 31+ 30+ 31+ 30+ 31+day;break;
case 9:j=31+ 28+ 31+ 30+ 31+ 30+ 31+ 31+ day;break;
case 10:j=31+ 28+ 31+ 30+ 31+ 30+ 31+ 31+ 30+day;break;
case 11:j=31+ 28+ 31+ 30+ 31+ 30+ 31+ 31+ 30+ 31+day;break;
case 12:j=31+ 28+ 31+ 30+ 31+ 30+ 31+ 31+ 30+ 31+ 30+ day;break;
}
if(i==1)j+=1;
printf("%d年%d月%d日是该年的第%d天",year,month,day,j);
return 0;
}
C语言入门
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