Codeforces Round #180 (Div. 2)-A. Snow Footprintsv

原题链接
A. Snow Footprints
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

Examples
input
9
..RRLL…
output
3 4
input
11
.RRRLLLLL..
output
7 5
Note
The first test sample is the one in the picture.
思路：由于题目中保证输入数据正确，所以我们发现不可能出现LR这种情况，所以只有可能的情况是全R或者全L或者n个R接n个L，所以只需要判断一下就行。

//http://codeforces.com/problemset/problem/298/A#include <algorithm>#include <iostream>#include <utility>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>using namespace std;typedef long long ll;const int MOD = int(1e9) + 7;//int MOD = 99990001;const int INF = 0x3f3f3f3f;const ll INFF = (~(0ULL)>>1);const double EPS = 1e-9;const double OO = 1e20;const double PI = acos(-1.0); //M_PI;const int fx[] = {-1, 1, 0, 0};const int fy[] = {0, 0, -1, 1};const int maxn=1000 + 5;char s[maxn];int a[maxn];int main(){    int n;    while(scanf("%d",&n)==1){        cin >> s;        int n=strlen(s);        memset(a,0,sizeof(a));        for(int i=0;i<n;i++){            if(s[i]=='R') a[i]=-1;            else if(s[i]=='L') a[i]=1;        }        int b,e;        for(int i=0;i<n;i++){            if(a[i]!=0){                b=i;                break;            }        }        for(int i=n-1;i>=0;i--){            if(a[i]!=0){                e=i;                break;            }        }        bool flag=false;//默认全是同一种颜色        int jie;//jie就是RL这个特征的部分的L的下标        for(int i=b+1;i<=e;i++){            if(a[i]!=a[b]){                flag=true;                jie=i;                break;            }        }        if(flag){            printf("%d %d\n",b+1,jie);        }        else{            if(a[b]==-1){//如果全是R                printf("%d %d\n",b+1,e+2);            }            else{//如果全是L                printf("%d %d\n",e+1,b);            }        }    }    return 0;}