Codeforces Round #180 (Div. 2) A. Snow Footprints 贪心

A. Snow Footprints

题目连接：

http://www.codeforces.com/contest/298/problem/A

Description

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.

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 Input

9

..RRLL...

Sample Output

3 4

Hint

题意

给你长度为n的字符，表示这是一条路径。

上面有很多脚印，R表示从这儿走到了右边，L表示从这儿走到了左边

问你根据脚印，判断起点和终点在哪儿。

题解：

最后图形肯定是L连成一堆，R连成一堆，起点就是开始的位置，终点就是L，R的交界处。

代码

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n;char s[5000];
    scanf("%d%s",&n,s);
    int len = strlen(s);
    int a,b;
    for(a=0;s[a]=='.';a++);
    for(b=a;s[a]==s[b];b++);
    if(s[a]=='R')
    {
        a++,b++;
        printf("%d %d\n",a,b);
    }
    else
    {
        printf("%d %d\n",b,a);
    }
}