Waterloo Cup: Basic practice of Special palindrome number

Basic Practice Special Palindrome number time limit: 1.0s memory Limit: 512.0MB problem description 123321 is a very special number, which reads from the left and reads the same from the right.
Enter a positive integer n, programming all such five-bit and six-bit decimal digits, satisfying the sum of the numbers equals N. Enter a row for the input format, including a positive integer n. The output format outputs an integer that satisfies the condition in a small to large order, and each integer occupies one row. Example input 52 sample output 899998
989989
998899 Data Scale and convention 1<=n<=54. Method of solving Problems: Set loop (10000->1000000), get the number of bits per digit, and compare the number of numbers entered. Converts a loop variable to a character array, reverse it, get a new string, and compare it to the original string (palindrome number if true). If the comparison results are true, the number of this palindrome is output.

Import Java.util.Scanner;

public class main{public

	static void Main (string[] args) {
		Scanner sc = new Scanner (system.in);
		int number = Sc.nextint ();
		int i,j,len,n;
		int sumt = 0;
		char []array = new char[6];
		String strorigin = "";
		String Strturn = "";
		StringBuilder Stbturn = new StringBuilder ();
		for (i = 10000 i < 1000000; i++) {
			n = i;
			Strorigin = integer.tostring (i);
			Array = Strorigin.tochararray ();
			while (n > 0) {
				sumt + = n%;
				n = n/10;
			}
			len = array.length;
			for (j = len-1 J >= 0; j--) {
				stbturn.append (array[j]);
			Strturn = Stbturn.tostring ();
			if (Strorigin.equals (Strturn)) {
				if (sumt = number) {
					System.out.println (i);
				}
			}
			Stbturn = new StringBuilder ();
			SUMT = 0;
		}
	}