OLI 课程 Java入学考试的五道题
Unit? 1::?Programming with Java
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機械、自動、不需要智慧地執行原始的內置指令。
字節碼相同,JVM不同(體現平臺)
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In modern computers instructions can be stored and manipulated as other types of data. 對,都是位運算
We cannot describe an algorithm without a programming language.錯,沒有語言也可以寫算法
compile:高級語言變成低級語言,設置一次就行,只翻譯 不執行 很快,形成字節碼。interpreter:要設置很多次,又翻譯又執行。
?z = x * y, 改y, z不變
用byte(帶符號整數)存100。Although such data values can be stored in bigger numeric data types, e.g. short, int, or long, the most appropriate is byte as it takes least memory and has the range from -128 to 127.
double可以是也可以不是浮點型小數?
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打印前100個prime numbers
第一個x設為2,然后逐個往后試驗
public static void main(String[] args) {Scanner scan = new Scanner(System.in);int N = scan.nextInt();int x = 2;for(int i = 0; i <= N; i++){int count = 0;for(int j = 1; j <= x; j++)if(x%j == 0)count++;if(count == 2)System.out.print(x + " ");x++;} }class driverFuction {
public static void main(String[] args)
{
Scanner scanner = new Scanner(System.in);
int x = 2;
int n = scanner.nextInt();
//count for n numbers
for(int i = 0; i <= n; i++) {
//reset count = 0;
int count = 0;
for (int j = 1; j <= x; j++) {
if (x % j == 0) count++;
}
if (count == 2) System.out.println("prime is x = " + x);
//add x
x++;
}
}
}
import java.util.Scanner;
public class ExArraySortElement {public static void main(String[] args) {int n, temp;//scanner class object creationScanner s = new Scanner(System.in);//input total number of elements to be readSystem.out.print("Enter the elements you want : ");n = s.nextInt();//integer array objectint a[] = new int[n];//read elements System.out.println("Enter all the elements:");for (int i = 0; i < n; i++) {a[i] = s.nextInt();//數組中的元素需要一個個地輸入
}//sorting elementsfor (int i = 0; i < n; i++) {for (int j = i + 1; j < n; j++) {if (a[i] > a[j]) {temp = a[i];a[i] = a[j];a[j] = temp;}}}//print sorted elements System.out.println("Ascending Order:");for (int i = 0; i < n ; i++) {System.out.println(a[i]);}} }
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// package whatever; // don't place package name!import java.io.*; import java.util.*; import java.lang.*;class driverFuction {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();//new an arrayint[] nums = new int[n];//input n numbersfor (int i = 0; i < n; i++) {a[i] = scanner.nextInt();}//sort for (int i = 0; i < n; i++) {for (int j = i + 1; j < n; j++) {if (nums[i] > nums[j]) {int temp = nums[i];nums[i] = nums[j];nums[j] = temp;}}}//outputfor (int i = 0; i < n; i++) {System.out.println("nums[i] = " + nums[i]);} } }https://www.includehelp.com/java-programs/sort-an-array-in-ascending-order.aspx
import java.util.Scanner; class PrimeCheck {public static void main(String args[]){ int temp;boolean isPrime=true;Scanner scan= new Scanner(System.in);System.out.println("Enter any number:");//capture the input in an integerint num=scan.nextInt();scan.close();for(int i=2;i<=num/2;i++){temp=num%i;if(temp==0){isPrime=false;break;}}//If isPrime is true then the number is prime else notif(isPrime)System.out.println(num + " is a Prime Number");elseSystem.out.println(num + " is not a Prime Number");} }
https://beginnersbook.com/2014/01/java-program-to-check-prime-number/
// package whatever; // don't place package name!
import java.io.*;
import java.util.*;
import java.lang.*;
class Palindrome
{
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
System.out.println("Enter a string to check if it is a palindrome");
String original = in.nextLine();
//int length = original.length();
String clean = original.replaceAll("\\s+", "").toLowerCase();
//單詞里面有空格,要用replaceall
int length = clean.length();
int forward = 0;
int backward = length - 1;
while (backward > forward) {
char forwardChar = clean.charAt(forward++);
char backwardChar = clean.charAt(backward--);
if (forwardChar != backwardChar)
System.out.println("false");
}
System.out.println("true");
}
}
import java.io.*;
import java.util.*;
import java.lang.*;
class Palindrome
{
public static void main(String args[])
{
String word, File file;
int count = 0;
Scanner scanner = new Scanner(file);
while (scanner.hasNextLine()) {
String nextToken = scanner.next();//表示下一行
if (nextToken.equalsIgnoreCase(word))
count++;
}
return count;
}
}
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轉載于:https://www.cnblogs.com/immiao0319/p/9164701.html
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