JAVA小题
1.根据天数（46）计算周数和剩余的天数2. 已知圆的半径radius= 1.5，求其面积3.判断用户输入的数是否是偶数4.实现计算器功能，用户输入两个数，并且当用户按1时：求出两个数的和按2时：求出两个数的差按3时：求出两个数的积4时：求出两个数的商否则，求出两个数的余
5.任意输入三个数，然后按从大到小的顺序输出6.任意输出五个数，判断里边有无重复的数7.用if…else if语句判断闰年问题8.要求输入一个代表年月日的8位整数，计算出其上一天和下一天是什么日期
1.根据天数（46）计算周数和剩余的天数
class day
{
private int day
;
public day()
{
day
=0;
}
public day(int day
)
{
this.day
=day
;
}
public void setDay(int day
)
{
this.day
=day
;
}
public int getDay()
{
return this.day
;
}
public int weak()
{
return day
/7;
}
public int residueDay()
{
return 365-day
;
}
}
public class theDay
{
public static void main(String args
[])
{
day Day
=new day(46);
System
.out
.println("天数"+Day
.getDay()+"是第"+Day
. weak()+"周,"+"剩余的天数是："+Day
.residueDay());
}
}
2. 已知圆的半径radius= 1.5，求其面积
import java
.lang
.Math
;
public class area
{
public static void main(String args
[])
{
double radius
= 1.50;
System
.out
.println("圆的面积是"+Math
.PI
*radius
*radius
);
}
}
3.判断用户输入的数是否是偶数
import
java
.util
.*
;
public class T1 {
public static void main(String
[] args
)
{
Scanner
input
=new Scanner(System
.in
);
int a
=input
.nextInt();
if(a
%2==0)
{
System
.out
.println("输入数是偶数");
}
else
System
.out
.println("输入数是奇数");
input
.close();
}
}
4.实现计算器功能，用户输入两个数，并且当用户
按1时：求出两个数的和
按2时：求出两个数的差
按3时：求出两个数的积
4时：求出两个数的商
否则，求出两个数的余
import
java
.util
.*
;
class caslputer{
private double x
;
private double y
;
public caslputer(){
x
=0.0;
y
=0.0;
}
public caslputer(double x
,double y
){
this.x
=x
;
this.y
=y
;
}
public void operation(int z
){
switch(z
){
case 1: System
.out
.println(x
+y
);break;
case 2: System
.out
.println(x
-y
);break;
case 3: System
.out
.println(x
*y
);;break;
case 4: System
.out
.println(x
/y
);break;
default:
System
.out
.println(x
%y
);
}}}
public class T1 {
public static void main(String
[] args
) {
Scanner
input
=new Scanner(System
.in
);
double a
=input
.nextDouble();
double b
=input
.nextDouble();
caslputer
x
=new caslputer(a
,b
);
int c
=input
.nextInt();
x
.operation(c
);
input
.close();
}
}
5.任意输入三个数，然后按从大到小的顺序输出
import java
.util
.*
;
public class T1 {
public static void main(String
[] args
) {
Scanner input
=new Scanner(System
.in
);
double a
=input
.nextDouble();
double b
=input
.nextDouble();
double c
=input
.nextDouble();
if(a
>b
)
if(b
>c
)
System
.out
.println(a
+" " + b
+" "+ c
);
else
{
if(a
>c
)
System
.out
.println(a
+" "+c
+" " +b
);
else
System
.out
.println(c
+" " +b
+" "+a
);
}
else
if(a
>c
)
System
.out
.println(b
+" " +a
+" "+c
);
else
{
if(b
>c
)
System
.out
.println(b
+" " +c
+" " +a
);
else
System
.out
.println(c
+" " +b
+" "+a
);
}
input
.close();
}
}
6.任意输出五个数，判断里边有无重复的数
import java
.util
.*
;
public class T1 {
public static void main(String
[] args
) {
Scanner input
=new Scanner(System
.in
);
double a
=input
.nextDouble();
double b
=input
.nextDouble();
double c
=input
.nextDouble();
double d
=input
.nextDouble();
double e
=input
.nextDouble();
int n
=0;
for(int i
=0;i
<2;i
++)
{
if((a
==b
)||(a
==c
)||(a
==d
)||(a
==e
))
break;
else if((b
==c
)||(b
==d
)||(b
==e
))
break;
else if((c
==d
)||(c
==e
))
break;
else if(d
==e
)
break;
n
=1;
}
if(n
==1)
System
.out
.println("无重复数据");
else
System
.out
.println("有重复数据");
input
.close();
}
}
7.用if…else if语句判断闰年问题
import java
.util
.*
;
public class T1 {
public static void main(String
[] args
) {
Scanner input
=new Scanner(System
.in
);
int year
=input
.nextInt();
if ((year
% 400 == 0) || ((year
% 100 == 0)
&& (year
% 4 == 0)))
System
.out
.println("这年是闰年！！！");
else
System
.out
.println("不是闰年！！！");
input
.close();
}
}
8.要求输入一个代表年月日的8位整数，计算出其上一天和下一天是什么日期
import java
.text
.DateFormat
;
import java
.util
.*
;
public class T1 {
public static void main(String
[] args
)
{
Scanner input
=new Scanner(System
.in
);
int a
=input
.nextInt();
int year
;
int month
;
int day
;
year
=a
/10000;
month
=a
/100%100;
day
=a
%100;
Calendar calendar
=Calendar
.getInstance();
calendar
.clear();
calendar
.set(year
, month
-1, day
-1);
DateFormat t
=DateFormat
.getDateInstance(DateFormat
.LONG
);
System
.out
.println("上一天"+t
.format(calendar
.getTime()));
calendar
.set(year
, month
-1, day
+1);
System
.out
.println("下一天"+t
.format(calendar
.getTime()));
input
.close();
}
}
