Problem:
Score : 400 points

Problem Statement

In this problem, we only consider strings consisting of lowercase English letters.

Strings s and t are said to be isomorphic when the following conditions are satisfied:

|s| = |t| holds.

For every pair i, j, one of the following holds:

s_i = s_j and t_i = t_j.

s_i \neq s_j and t_i \neq t_j.

For example, abcac and zyxzx are isomorphic, while abcac and ppppp are not.

A string s is said to be in normal form when the following condition is satisfied:

For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison.

For example, abcac is in normal form, but zyxzx is not since it is isomorphic to abcac, which is lexicographically smaller than zyxzx.

You are given an integer N.
Print all strings of length N that are in normal form, in lexicographically ascending order.

Constraints

1 \leq N \leq 10

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N

Output

Assume that there are K strings of length N that are in normal form: w_1, \ldots, w_K in lexicographical order.
Output should be in the following format:

w_1
:
w_K

Sample Input 1

1

Sample Output 1

a

Sample Input 2

2

Sample Output 2

aa
ab

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)



import java.io.IOException;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) throws IOException {

		Scanner scan = new Scanner(System.in);
		int n;
		n = scan.nextInt();

		String a = "a";
		String b = "b";

		if (n == 1) {
			System.out.println(a);
			return;
		}

		byte output = 0b0000;

		while (true) {
			//bit値を文字列型に置換
			String ans = String.valueOf(output);
			ans = ans.replace("0", "a");
			ans = ans.replace("1", "b");
			if (ans.charAt(0) == 'b') {
				return;
			}
			System.out.println(ans);

			//bit値を1進める
			output = 0b1;
		}
	}

	public static long lessThanSqrt(long a) {
		return longSqrt(a - 1);
	}

	public static long longSqrt(long a) {
		long b = (long) Math.sqrt(a);
		// 得られた値を2乗して元の値を超える場合は
		// 誤差で1大きくなっているため
		// 誤差修正のため1引いた値を返す
		if (b * b > a) {
			b--;
		}
		return b;
	}

	/**
	 * 余りを求める。
	 * @param a
	 * @param b
	 * @return 余り。余りがマイナスの時はプラスで返す。
	 */
	public static long myMod(long a, long b) {
		long ret = 0;

		ret = a % b;

		if (ret < 0) {
			ret += b;
		}

		return ret;
	}

	/**
	 * 乗数を求める ※mod対応
	 * @param kisu 基数
	 * @param josu 乗数
	 * @param primeNumMod_0 法 mod値 使わない場合は「0」
	 * @return
	 */
	public static long myPowWithMod(long kisu, long josu, long primeNumMod_0) {

		long ans = 1;

		//乗数が0の場合→1
		if (josu == 0)
			return 1;

		if (primeNumMod_0 == 0) {
			//modを使わないとき

			while (josu > 0) {

				if (josu % 2 == 0) {
					//累乗が偶数の時

					kisu *= kisu;
					josu = josu / 2;
				} else {
					//累乗が奇数の時
					ans *= kisu;
					josu--;
				}
			}

			return ans;

		} else {
			//modを使うとき

			//初めに桁を落とす
			kisu = myMod(kisu, primeNumMod_0);

			while (josu > 0) {

				if (josu % 2 == 0) {
					//累乗が偶数の時

					kisu *= myMod(kisu, primeNumMod_0);
					josu = josu / 2;

				} else {
					//累乗が奇数の時
					ans *= myMod(kisu, primeNumMod_0);
					josu--;

				}
			}

			return myMod(ans, primeNumMod_0);

		}

	}

}