1. Known March 17, 2019 is a Sunday, to obtain user input 8-bit integer that represents the date of the week 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) ;
DateFormat t= DateFormat. getDateInstance ( DateFormat. FULL) ;
System. out. println ( t. format ( calendar. getTime ( ) ) ) ;
input. close ( ) ;
}
} 2. obtain all even and less than 100 public class T1 {
public static void main ( String[ ] args) {
int s= 0 ;
for ( int i= 0 ; i<= 100 ; i++ )
{
if ( i% 2 == 0 )
s= s+ i;
if ( i== 100 )
System. out. println ( s) ;
}
}
} 3. do-while implementation: Output Celsius and Fahrenheit table, which requires a temperature of from 0 degrees to 250 degrees Celsius, 20 degrees every one, the entry in the table is not more than 10. Conversion relationship: Fahrenheit Celsius * = 9/32 + 5.0; Tip: 1, cyclic operation: calculation Celsius, and outputs the control entry 2 cycling conditions: Entry <= 10 && Celsius <= 250 public class T1 {
public static void main ( String[ ] args) {
double fahrenheit;
int degreeCelsius= 0 ;
int x= 1 ;
do
{
if ( ( degreeCelsius% 20 == 0 ) )
{
fahrenheit= degreeCelsius* 9 / 5.0 + 32 ;
x++ ;
System. out. println ( "华氏温度:" + fahrenheit+ "摄氏温度 :" + degreeCelsius) ;
}
degreeCelsius++ ;
} while ( ( x<= 10 ) && ( degreeCelsius<= 250 ) ) ;
}
}
4. The output 100 can be less than the number of the front 57 of the divisible public class T1 {
public static void main ( String[ ] args)
{ int c= 0 ;
for ( int i= 0 ; i< 100 ; i++ )
{
if ( i% 7 == 0 )
{ c++ ;
if ( c<= 5 )
{
System. out. println ( i) ;
}
}
}
}
} The cylinder house a total of 50 liters. Existing 15 liters. Each can pick five liters. To pick a few times to pick over. public class T1 {
public static void main ( String[ ] args)
{ int water= 50 ;
int nowWater= 15 ;
int spurWater= 5 ;
int x;
x= ( water- nowWater) / spurWater;
System. out. println ( "要挑" + x+ "次才能挑满" ) ;
}
} 6. obtain 1! +2! + ... + n! How much? (Use while to do) import
java. util. *;
public class T1 {
public static void main ( String[ ] args)
{
int s= 1 ;
int i= 1 ;
int a= 0 ;
Scanner
input= new Scanner ( System. in) ;
System. out. println ( "请输入n:" ) ;
int n= input. nextInt ( ) ;
while ( i<= n)
{
s= s* i;
i++ ;
a= s+ a;
}
System. out. println ( a) ;
input. close ( ) ;
}
} 7. require users to enter a number, use the do-while it reversed in the past! import java. util. *;
public class T1 {
public static void main ( String[ ] args)
{
Scanner input= new Scanner ( System. in) ;
String a= input. next ( ) ;
System. out. println ( new StringBuilder ( a) . reverse ( ) . toString ( ) ) ;
input. close ( ) ;
}
} 8 ... 1000 is obtained within all divisible by 4 and 5 and can not be divisible by 3 and the number of public class T1 {
public static void main ( String[ ] args)
{
int c= 0 ;
int a= 0 ;
for ( int i= 1 ; i<= 1000 ; i++ )
{
if ( i% 3 != 0 )
if ( ( i% 4 == 0 ) && ( i% 5 == 0 ) )
{
c= i;
a= c+ a;
if ( i== 1000 )
System. out. println ( a) ;
}
}
}
}
