A. Partial Teacher
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

A teacher decides to give toffees to his students. He asks n students to stand in a queue. Since the teacher is very partial, he follows the following rule to distribute toffees.

He looks at the first two students and gives more toffees to the student having higher marks than the other one. If they have the same marks they get the same number of toffees. The same procedure is followed for each pair of adjacent students starting from the first one to the last one.

It is given that each student receives at least one toffee. You have to find the number of toffees given to each student by the teacher such that the total number of toffees is minimum.

Input

The first line of input contains the number of students n (2 ≤ n ≤ 1000). The second line gives (n - 1) characters consisting of "L", "R" and "=". For each pair of adjacent students "L" means that the left student has higher marks, "R" means that the right student has higher marks and "=" means that both have equal marks.

Output

Output consists of n integers separated by a space representing the number of toffees each student receives in the queue starting from the first one to the last one.

Examples
input 
5
LRLR
output 
2 1 2 1 2
input 
5
=RRR
output 
1 1 2 3 4

题意：n个学生排一列一行，老师给学生发奶糖，已知n-1个比较关系（相邻两个学生奶糖数量的大小关系）

求老师最少发几颗？


思路：贪心+DP+模拟，遇到L，那么当前变1，再把变1对之前的影响反推

#include <bits/stdc++.h>

typedef long long ll;
using namespace std;

const int N=1005;
char str[N];
int dp[1005];

int main(void){
    int n;
    cin >> n;
    cin >> str+1;
    dp[0]=1;
    for(int i=1;i<=n-1;i++){
        if(str[i]=='=') dp[i]=dp[i-1];
        else if(str[i]=='R')    dp[i]=dp[i-1]+1;
        else{
            dp[i]=1;
            if(dp[i-1]==1){
                dp[i-1]=2;
                for(int j=i-1;j>=1;j--){
                    if(str[j]=='=') dp[j-1]=dp[j];
                    else if(str[j]=='R') break;
                    else if(str[j]=='L'){
                        if(dp[j-1]==dp[j])  dp[j-1]++;
                    }
                }
            }
        }
    }
    for(int i=0;i<n;i++)
        printf("%d ",dp[i]);
    return 0;
}