codeforce 298 A Snow Footprints

最新推荐文章于 2021-02-17 00:47:56 发布

原创最新推荐文章于 2021-02-17 00:47:56 发布 · 1.1k 阅读

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#水题

本文探讨了一种通过分析雪地足迹来推断行走路径的方法，详细阐述了路径长度、足迹类型以及覆盖机制对路径推断的影响。通过实例分析，展示了如何利用输入的足迹描述来确定起点和终点位置。

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 thet-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.

#include <stdlib.h>
#include <string.h>
#include <stdio.h>
char input[1010];
int r1=0,r2=0,l1=0,l2=0;
void getr()
{
int i;
for(i=0;i<strlen(input);i++)
{
if(input[i]=='R')
{
r1=i;
break;
}
}
if(i==strlen(input))
{
r1=-1;
r2=-1;
return;
}
else
{
for(i=strlen(input)-1;i>=0;i--)
{
if(input[i]=='R')
{
r2=i;
break;
}
}
}
}
void getl()
{
int i;
for(i=strlen(input)-1;i>=0;i--)
{
if(input[i]=='L')
{
l2=i;
break;
}
}
if(i<0)
{
l1=-1;
l2=-1;
return;
}
else
{
for(i=0;i<strlen(input);i++)
{
if(input[i]=='L')
{
l1=i;
break;
}
}
}
}
void solve()
{
if(r1==-1)
{
printf("%d %d\n",l2+1,l1);
}
else if(l2==-1)
{
printf("%d %d\n",r1+1,r2+2);
}
else
{
printf("%d %d\n",r1+1,r2+1);
}
}
int main(int argc, char *argv[])
{
//freopen("data.in","r",stdin);
int len;
while(scanf("%d",&len)!=EOF)
{
scanf("%s",input);
getr();
getl();
solve();
}
return 0;
}