英文:
Java loop vertical
问题
import java.util.Scanner;public class Main {public static void main(String args[]) {Scanner in = new Scanner(System.in);int num = in.nextInt();int rev = 0;while (num != 0) {rev = rev * 10;rev = rev + num % 10;num = num / 10;}while (rev != 0) {System.out.println(rev % 10);rev = rev / 10;}}}
英文:
My code basically arranged the number into reverse order for example 415 the program will arrange it into 514 well my code is correct but I have a problem the output should be vertical.
> expected output
> 5
> 1
> 4
import java.util.Scanner;public class Main{public static void main(String args[]){Scanner in = new Scanner(System.in);int num = in.nextInt();int rev=0;while( num != 0 ){rev = rev * 10;rev = rev + num%10;num = num/10;}System.out.println(rev);}}
答案1
得分: 5
你只需要像这样做：
while (num != 0){System.out.println(num % 10);num = num / 10;}
英文:
You just need to do it like this:
while( num != 0 ){System.out.println(num % 10);num = num / 10;}
答案2
得分: 0
这是另一种方法，您可以使用此方法获取超过整型数据类型所能容纳的过大的整数。
Scanner in = new Scanner(System.in);String s = in.nextLine();for (int i = s.length() - 1; i >= 0; i--) {System.out.println(s.charAt(i));}
英文:
I am leaving this as an alternate method. By this you can take int which are are too big to fit in int datatype
Scanner in = new Scanner(System.in);String s = in.nextLine();for (int i = s.length() - 1; i >= 0; i--) {System.out.println(s.charAt(i));}
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