Codeforces-298a G Snow Footprints

最新推荐文章于 2019-07-01 21:58:42 发布

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本文介绍了一个关于雪地足迹的编程挑战问题。任务是通过分析雪地上留下的左右足迹来确定一只北极熊的起点和终点。文章提供了一个简单的模拟解决方案，并附带了示例输入输出，帮助理解如何根据给定的足迹字符串确定移动路径。

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There is a straight snowy road, divided into n blocks. The blocks are numbered from 1 to n from left to right. If one moves from the i-th block to the (i + 1)-th block, he will leave a right footprint on the i-th block. Similarly, if one moves from the i-th block to the (i - 1)-th block, he will leave a left footprint on the i-th block. If there already is a footprint on the i-th block, the new footprint will cover the old one.

$\text{[math]}$

At the beginning, there were no footprints. Then polar bear Alice starts from the s-th block, makes a sequence of moves and ends in the t-th block. It is known that Alice never moves outside of the road.

You are given the description of Alice's footprints. Your task is to find a pair of possible values of s, t by looking at the footprints.

Input

The first line of the input contains integer n (3 ≤ n ≤ 1000).

The second line contains the description of the road — the string that consists of n characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint).

It's guaranteed that the given string contains at least one character not equal to ".". Also, the first and the last character will always be ".". It's guaranteed that a solution exists.

Output

Print two space-separated integers — the values of s and t. If there are several possible solutions you can print any of them.

Sample test(s)

Input

9
..RRLL...

Output

3 4

Input

11
.RRRLLLLL..

Output

7 5

Note

The first test sample is the one in the picture.

————————————————————水水的分割线————————————————————

思路：简单的模拟。考特判能力。因为只需要打印任意一个解，稍加思考就可以得到一个结论：一般情况下，第一个字母如果是R，那么它就是起点。而终点就是最后一个R后面的点。

第一个字母如果是L，起点就是最后一个L后面的点，起点就是第一个L前面的点。

代码如下：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <memory.h>
#include <stdlib.h>
int main(){
	char foot[1010];
	int n, c, i = 0, left = 0;
	int s, t;
	scanf("%d", &n);getchar();
	for(;;){
        c = getchar();
        left++;
        if(c != '.')
            break;
	}
	for(;;){
        foot[i++] = c;
        c = getchar();
        if(c == '.')
            break;
	}
	n = i;
//printf("n is %d now\n", n);
    if(foot[0] == 'L'){
        s = n-1;
        t = -1;
    }
    if(foot[0] == 'R'){
        s = 0;
        for(i = 0;;i++)
            if(foot[i] != 'R')
                break;
        t = i-1;
        if(i == n)  t = i;
    }
    printf("%d %d\n", s+left, t+left);
	return 0;
}