Codeforces 3A. Shortest path of the king

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本文介绍了一个基于Codeforces3A的编程题解决方案，任务是在8*8棋盘中找到两格之间的最短移动路径，利用简单的数学计算确定移动方向。

【题目来源】：Codeforces 3A.

A. Shortest path of the king

The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square t. As the king is not in habit of wasting his time, he wants to get from his current position s to square t in the least number of moves. Help him to do this.

In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).

Input

The first line contains the chessboard coordinates of square s, the second line — of square t.

Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1to 8.

Output

In the first line print n — minimum number of the king's moves. Then in n lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.

L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.

Examples

input

a8
h1

output

7
RD
RD
RD
RD
RD
RD
RD

【题目大意】：在8*8的棋盘上任选两个点，找到两个点之间横纵距离的最大值，并输出从第一个点到第二个点的最短路径，共有上，下，左，右，左上，左下，右上，右下，八个方向可以走。

【AC代码】：

import java.util.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s1 = sc.next();
		String s2 = sc.next();
		char s11 = s1.charAt(0);
		char s12 = s1.charAt(1);
		char s21 = s2.charAt(0);
		char s22 = s2.charAt(1);
		System.out.println(Math.max(Math.abs((int) (s11 - s21)),
				Math.abs((int) (s12 - s22))));
		while (s11 != s21 || s12 != s22) {
			if (s11 < s21) {
				s11++;
				System.out.print("R");
			} else if (s11 > s21) {
				s11--;
				System.out.print("L");
			}
			if (s12 > s22) {
				s12--;
				System.out.print("D");
			} else if (s12 < s22) {
				s12++;
				System.out.print("U");
			}
			System.out.println();
		}
	}
}