CodeForces 298A　Snow Footprints

最新推荐文章于 2017-07-29 23:55:00 发布

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最新推荐文章于 2017-07-29 23:55:00 发布

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本文提供了一个解决SnowFootprints问题的代码实现，详细解释了如何通过字符串操作找到小球的起始位置和结束位置。代码示例涵盖了输入处理、逻辑判断以及输出结果的生成。

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1、http://codeforces.com/problemset/problem/298/A

2、题目大意：给定一个小球，可以往左往右移动，如果小球从第i个位置移到第i+1个位置，那么第i个位置将置为R，如果从第i个位置移到第i-1个位置，那么第i个位置将置为Ｌ，现在已知最终留下的ＬＲ序列，求出小球的起始位置

３、题目：

A - Snow Footprints

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 298A

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.

$\text{[math]}$

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 Input

Input

9
..RRLL...

Output

3 4

Input

11
.RRRLLLLL..

Output

7 5

Hint

The first test sample is the one in the picture.

４、代码：

#include<stdio.h>
#include<string.h>
char str[1005];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%s",str);
        int len=strlen(str);
        int flag=0,flag1=0;
        int start,end;
        for(int i=0;i<len;i++)
        {
            if(str[i]=='R'&&flag==0)
            {
                flag=1;
                start=i;
                end=i;
            }
            if(str[i]=='R'&&str[i+1]=='L')
            {
                end=i;
                break;
            }
            if(str[i]=='R' && str[i+1]=='.')
            {
                end=i+1;
                break;
            }
            if(str[i]=='L'&&flag1==0)
            {
                flag1=1;
                start=i;
                end=i-1;
            }
            if(str[i]=='L'&&str[i+1]=='R')
            {
                start=i;
            }
            if(str[i]=='L' && str[i+1]=='.')
            start=i;
        }
        printf("%d %d\n",start+1,end+1);
    }

    return 0;
}