1.99 multiplication table
public class 九九乘法表 {
public static void main ( String[ ] args) {
System. out. println ( "方法一:" ) ;
for ( int i= 1 ; i<= 9 ; i++ ) {
for ( int j= 1 ; j<= i; j++ ) {
System. out. print ( i+ "*" + j+ "=" + i* j+ "\t" ) ;
}
System. out. println ( ) ;
}
System. out. println ( "=================================================================================" ) ;
System. out. println ( "方法二:" ) ;
int i= 1 ;
while ( i<= 9 ) {
int j= 1 ;
while ( j<= i) {
System. out. print ( i+ "*" + j+ "=" + i* j+ "\t" ) ;
j++ ;
}
System. out. println ( ) ;
i++ ;
}
}
}
2. Define a fixed-length array with a length of 100. Each element is a randomly generated number between 0 and 9. Count the number of occurrences of each number.
import java. util. Arrays;
public class 获取随机数组中每个元素出现的次数 {
public static void main ( String[ ] args) {
java. util. Random Random= new java. util. Random ( ) ;
int [ ] array= new int [ 100 ] ;
for ( int i= 0 ; i< array. length; i++ ) {
array[ i] = Random. nextInt ( 10 ) ;
}
System. out. println ( "生成的数组是:" ) ;
System. out. println ( Arrays. toString ( array) ) ;
int [ ] times= new int [ array. length] ;
for ( int i= 0 ; i< 10 ; i++ ) {
for ( int j= 0 ; j< array. length; j++ ) {
if ( i== array[ j] ) {
times[ i] ++ ;
}
}
}
for ( int i= 0 ; i< 10 ; i++ ) {
System. out. println ( i+ "出现了" + times[ i] + "次!" ) ;
}
}
}
3. Count prime numbers within 1 ~ 1000
public class 统计质数 {
public static void main ( String[ ] args) {
System. out. println ( "1至1000以内的质数如下:" ) ;
System. out. print ( 2 + "\t" ) ;
int time= 1 , flag= 1 ;
OUT:
for ( int i= 3 ; i<= 1000 ; i+= 2 ) {
for ( int j= 2 ; j< i; j++ ) {
if ( i% j== 0 ) {
continue OUT;
}
}
if ( flag% 30 == 0 ) {
System. out. println ( ) ;
flag= 0 ;
}
time++ ;
flag++ ;
System. out. print ( i+ "\t" ) ;
}
System. out. println ( ) ;
System. out. println ( "质数总个数为:" + time) ;
}
}
4. Input value sorting: (bubble sorting method)
import java. util. Arrays;
public class 排序 {
public static void main ( String[ ] args) {
java. util. Scanner s= new java. util. Scanner ( System. in) ;
System. out. println ( "输入需要排序数组的个数:" ) ;
int x= s. nextInt ( ) ;
int n[ ] = new int [ x] ;
System. out. println ( "输入需要排序的数:" ) ;
for ( int i= 0 ; i< x; i++ ) {
n[ i] = s. nextInt ( ) ;
} s. close ( ) ;
System. out. println ( "排序前:" ) ;
System. out. println ( Arrays. toString ( n) ) ;
for ( int i= 0 ; i< n. length- 1 ; i++ ) {
for ( int j= 0 ; j< n. length- 1 - i; j++ ) {
if ( n[ j] > n[ j+ 1 ] ) {
int tmp= n[ j] ;
n[ j] = n[ j+ 1 ] ;
n[ j+ 1 ] = tmp;
}
}
}
System. out. println ( "=====================================================================" ) ;
System. out. println ( "排序后:" ) ;
System. out. println ( Arrays. toString ( n) ) ;
}
}
5. Given a natural number n, find the number between 1 and n to form a non-repetitive three-digit number
import java. util. Scanner;
public class 不同数的组合 {
public static void main ( String[ ] args) {
Scanner s= new Scanner ( System. in) ;
int n= s. nextInt ( ) ;
s. close ( ) ;
int num= 0 ;
for ( int i= 1 ; i<= n; i++ ) {
for ( int j= 1 ; j<= n; j++ ) {
for ( int k= 1 ; k<= n; k++ ) {
if ( i!= j&& j!= k&& k!= i) {
int date= i* 100 + j* 10 + k;
System. out. println ( date+ " " ) ;
num++ ;
}
}
}
}
System. out. println ( "总个数:" + num) ;
}
}
6. Question 5 expansion (array version): first input an n, then give n numbers, find all three digits composed of these n numbers randomly
import java. util. Scanner;
public class 不同数的组合数组版 {
public static void main ( String[ ] args) {
Scanner s= new Scanner ( System. in) ;
int n= s. nextInt ( ) ;
int [ ] a= new int [ n] ;
for ( int i= 0 ; i< a. length; i++ ) {
a[ i] = s. nextInt ( ) ;
}
s. close ( ) ;
int num= 0 ;
for ( int i= 0 ; i< a. length; i++ ) {
for ( int j= 0 ; j< a. length; j++ ) {
for ( int k= 0 ; k< a. length; k++ ) {
if ( a[ i] != a[ j] && a[ j] != a[ k] && a[ k] != a[ i] ) {
int date= a[ i] * 100 + a[ j] * 10 + a[ k] ;
System. out. println ( date+ " " ) ;
num++ ;
}
}
}
}
System. out. println ( "总个数:" + num) ;
}
}
7. Randomly generate 10 non-repeating numbers between 1 and 20 (excluding 20), and output the sorted array: from small to large; from large to small;
Bubble sorting method:
import java. util. Arrays;
import java. util. Random;
public class 冒泡排序 {
public static void main ( String[ ] args) {
Random Ran= new Random ( ) ;
int [ ] a= new int [ 10 ] ;
OUT:
for ( int i= 0 ; i< a. length; i++ ) {
int date= Ran. nextInt ( 20 ) ;
for ( int j= 0 ; j< a. length; j++ ) {
if ( a[ j] == date) {
i-- ;
continue OUT;
}
}
a[ i] = date;
}
System. out. println ( "排序前数组中的信息:" + Arrays. toString ( a) ) ;
for ( int i= 0 ; i< a. length- 1 ; i++ ) {
for ( int j= 0 ; j< a. length- 1 - i; j++ ) {
if ( a[ j] > a[ j+ 1 ] ) {
int temp= a[ j] ;
a[ j] = a[ j+ 1 ] ;
a[ j+ 1 ] = temp;
}
}
}
System. out. println ( "由小往大排序后数组中的信息:" + Arrays. toString ( a) ) ;
for ( int i= 0 ; i< a. length- 1 ; i++ ) {
for ( int j= 0 ; j< a. length- 1 - i; j++ ) {
if ( a[ j] < a[ j+ 1 ] ) {
int temp= a[ j] ;
a[ j] = a[ j+ 1 ] ;
a[ j+ 1 ] = temp;
}
}
}
System. out. println ( "由大往小排序后数组中的信息:" + Arrays. toString ( a) ) ;
}
}