C语言程序设计:输入某年某月输出某月有多少天？

Posted 2023-04-08

#include <stdio.h>
void main()

int year ,month;
int month_day(int year,int month);
printf("输入年份：");
scanf("%d",&year);
printf("输入月份：");
scanf("%d",&month);
printf("%d年%d月有%d天",year,month,month_day(year,month));


int month_day(int year,int month)

int day;
switch(month)

int day;

case 1:
case 3:
case 5:
case 7:
case 8:
case10:
case 12:
day = 31;

break;
case 2:
if(year%100!=0&&year%4==0)
day = 29;
else

day = 28;
break;
case 4:
case 6:
case 9:
case 11:
day = 30;
break;
default:
printf("输入月份不存在")；

return day;

参考技术A #include <stdio.h>
int main ()

int year,month,day;
day=0;
printf("请输入年份和月份：\n");
scanf("%d %d",&year,&month);
switch(month)

case 1 :
case 3 :
case 5 :
case 7 :
case 8 :
case 10 :
case 12 :
day=31; break;
case 4 :
case 6 :
case 9 :
case 11 :
day=30; break;
case 2 :
if(year%4==0&&year%100!=0 || year%400==0)
day=29;
else day=28; break;
default:
printf("Date error!"); break;

printf("day=%d\n",day);
return 0;
参考技术B #include
<stdio.h>
void
main()

int
year
,month;
int
month_day(int
year,int
month);
printf("输入年份：");
scanf("%d",&year);
printf("输入月份：");
scanf("%d",&month);
printf("%d年%d月有%d天",year,month,month_day(year,month));

int
month_day(int
year,int
month)

int
day;
switch(month)

int
day;
case
1:
case
3:
case
5:
case
7:
case
8:
case10:
case
12:
day
=
31;
break;
case
2:
if(year%100!=0&&year%4==0)
day
=
29;
else
day
=
28;
break;
case
4:
case
6:
case
9:
case
11:
day
=
30;
break;
default:
printf("输入月份不存在")；

return
day;
参考技术C 用数组和条件语句做

1097某年某月天数

描述

打印某年某月有多少天。

 

输入

输入一行，包含2个整数 分别代表年和月

 

输出

输出一行，包含1个整数,表示该年的这个月份一共有多少天.

 

输入样例 1 

2008 2

输出样例 1

29

提示

闰年的计算方法：
1.年数能被4整除，并且不能被100整除；
2.能被400整除的整数年份。

 

错误示范：

 1 #include<iostream>
 2 #include<iomanip>
 3 using namespace std;
 4 int main()
 5 {
 6     int a;
 7     int x;
 8     cin>>a>>x;
 9     if(a%4==0&&a%100!=0||a%400==0)
10     {
11         if(x==2) cout<<"29";
12     }
13     else 
14     {
15         switch(x)
16         {
17             case 1:
18                 cout<<"31";
19                 break;
20             case 2:
21                 cout<<"29"; break;
22             case 3:
23                 cout<<"31"; break;
24             case 4:
25                 cout<<"30"; break;
26             case 5:
27                 cout<<"31"; break;
28             case 6:
29                 cout<<"30"; break;
30             case 7:
31                 cout<<"31"; break;
32             case 8:
33                 cout<<"31"; break;
34             case 9:
35                 cout<<"30"; break;
36             case 10:
37                 cout<<"31"; break;
38             case 11:
39                 cout<<"30"; break;
40             case 12:
41                 cout<<"31"; break;
42         }
43     }
44     return 0;
45 }

 

这题不难，但是对闰年的考虑加大了难度。

考虑到二月二十九号的情况题目就变得简单了。

正确代码：

 1 #include<iostream>
 2 #include<iomanip>
 3 using namespace std;
 4 int main()
 5 {
 6     int a;
 7     int x;
 8     cin>>a>>x;
 9     if(a%4==0&&a%100!=0||a%400==0)
10     {
11         if(x==2) cout<<"29";
12     }
13     else 
14     {
15         switch(x)
16         {
17             case 1:
18                 cout<<"31";
19                 break;
20             case 2:
21                 cout<<"28"; break;
22             case 3:
23                 cout<<"31"; break;
24             case 4:
25                 cout<<"30"; break;
26             case 5:
27                 cout<<"31"; break;
28             case 6:
29                 cout<<"30"; break;
30             case 7:
31                 cout<<"31"; break;
32             case 8:
33                 cout<<"31"; break;
34             case 9:
35                 cout<<"30"; break;
36             case 10:
37                 cout<<"31"; break;
38             case 11:
39                 cout<<"30"; break;
40             case 12:
41                 cout<<"31"; break;
42         }
43     }
44     return 0;
45 }