文章标题 CoderForces 298A： Snow Footprints（水）

Snow Footprints

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

Input

9

..RRLL…

Output

3 4

Input

11

.RRRLLLLL..

Output

7 5

Hint

The first test sample is the one in the picture.

题意：有n个格子，有一个人每走过这些格子，这个格子就会留下脚印，往右走就留下右边的脚印，左走就留下左边的脚印，然后现在有一些脚印，当再一次走到同一个格子，会把原来的脚印给覆盖掉，要我们求出这个人的起始点和终点

分析：首先有三种情况，有左边和右边的脚印，此时就只需要记住左边脚印（L）的第一个脚印的位置还有右边脚印（R）最右边的位置；另外两种情况就是全都是（R）或者都是（L）

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue>
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
int n;
int main ()
{
string a;
while (scanf ("%d",&n)!=EOF){
cin>>a;
int len = a.length();
int r_first=0,r_end=len;
int l_first=0,l_end=len;
for (int i=1;i<len;i++){
//记录第一个和最后一个的位置
if (a[i]=='R'&&a[i-1]!='R'){
r_first=i+1;
}
if (a[i]=='R'&&a[i+1]!='R'){
r_end=i+1;
}
if (a[i]=='L'&&a[i-1]!='L'){
l_first=i+1;
}
if (a[i]=='L'&&a[i+1]!='L'){
l_end=i+1;
}
}
if (r_end!=len&&l_first!=0){
//两种脚印都有
printf ("%d %d\n",l_first,r_end+1);
}
if (r_first==0){
//只有左脚印
printf ("%d %d\n",l_end,l_first-1);
}
if(l_first==0){
//只有右脚印
printf ("%d %d\n",r_first,r_end+1);
}
}
return 0;
}