Gym - 102219A：Mental Rotation（模拟）

http://codeforces.com/gym/102219/problem/A

Mental rotation is a difficult thing to master. Mental rotation is the ability to imagine in your mind how an object would look like from a viewer's side if you were to rotate it in a particular direction. It is something very important for engineers to learn and master, especially if you are required to do engineering drawing. There are many stages to these mental rotation tasks. We will approach a simple rotation task in this problem.

If you are given the following square -

After rotating it to the right once, it will look like the following -

After rotating it to the left once, it will look like the following -

For this problem you will be given an initial square of size n and a list of rotations to perform.

Your task is to output the final representation of the square after performing all the rotation.

Input

The first line of input contains an integer N(1 ≤ N ≤ 1000). and a string containing a combination of ′L′ and ′R′. Where ′L′ means left rotation and ′R′ means right rotation. Length of the string will not exceed 100. Starting from the second line, the following N line each will contain N of the following characters (>, <, ∨, ∧ or .). Empty space is indicated by a '.' (dot).

Output

The output should be N

lines, each containing N characters representing the final representation of the square after all the rotation. For ∨ and ∧ use v

(small v) and Shift+6 respectively.

Examples

Input

3 R
>v>
...
<^<

Output

^.v
>.<
^.v

Input

3 L
>v>
...
<^<

Output

^.v
>.<
^.v

Input

3 LL
>v>
...
<^<

Output

>v>
...
<^<

题意分析：

给出一个矩阵，在给出一个字符串，L 代表逆时针， R 代表顺时针，求旋转后的矩阵，旋转后相应的字符也会转变，如v顺时针旋转是<,逆时针旋转是>。

解题思路：

先把顺时针旋转次数和逆时针旋转次数记录，转4次之后会回到原位，所以都可以对4取余，逆时针1次=顺时针3次，所以又可以全部转化成顺时针旋转。

旋转我先用一个数组存旋转后的矩阵，再把矩阵复制到原数组中，把第一排旋转到倒数第一列，把第二排旋转到倒数第二列，以此类推。

#include <stdio.h>
#include <string.h>
#define N 1020
char a[N][N], a1[N][N], str[N];
int n;
void Right()
{
	for(int i=1; i<=n; i++)
		for(int j=1; j<=n; j++)
		{
			if(a[i][j]=='^')
				a1[j][n-i+1]='>';
			else if(a[i][j]=='v')
				a1[j][n-i+1]='<';
			else if(a[i][j]=='<')
				a1[j][n-i+1]='^';
			else if(a[i][j]=='>')
				a1[j][n-i+1]='v';
			else a1[j][n-i+1] ='.';
		}
	for(int i=1; i<=n; i++)
		for(int j=1; j<=n; j++)
			a[i][j]=a1[i][j];
}
int main()
{
	int l, r;
	while(scanf("%d%s", &n, str)!=EOF)
	{
		for(int i=1; i<=n; i++)
			scanf("%s", a[i]+1);
		l=0;r=0;
		for(int i=0; i<strlen(str); i++)
		{
			if(str[i]=='L')
				l++;
			else
				r++;
		}
		l%=4;
		r%=4;
		r=(r-l+4)%4;
		while(r--)
			Right();
		for(int i=1; i<=n; i++)
			printf("%s\n", a[i]+1);
	}
	return 0;
}