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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