Codeforces 67A Partial Teacher【dp】

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

5LRLR

Output

2 1 2 1 2

Input

5=RRR

Output

1 1 2 3 4

题目大意：

一共有N个人站成一排，对应每相邻的两个人都能确定两个人分到的糖果数谁更多一些（当然也可能相等）

保证没人必须至少分配一个糖的情况下，怎样分配能够满足条件并且分配的糖果总数最少。

思路：

1、设定dp【i】表示第i个人拿到的糖果数。

2、那么对应有三种状态转移方式：

①如果当前符号是=：dp【i】=dp【i-1】；

②如果当前符号是R：dp【i】=dp【i-1】+1；

③如果当前符号是L：dp【i】=1；

此时如果dp【i-1】也是1，那么对应向前推进，dp【i-1】+1；

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;int dp[1505];char a[1515];int main(){    int n;    while(~scanf("%d",&n))    {        dp[0]=1;        scanf("%s",a+1);        for(int i=1;i<=n-1;i++)        {            if(a[i]=='=')dp[i]=dp[i-1];            else if(a[i]=='R')dp[i]=dp[i-1]+1;            else            {                dp[i]=1;                if(dp[i-1]==1)                {                    dp[i-1]++;                    for(int j=i-1;j>=1;j--)                    {                        if(a[j]=='=')dp[j-1]=dp[j];                        else if(a[j]=='L'&&dp[j-1]==dp[j])dp[j-1]++;                        else break;                    }                }            }        }        for(int i=0;i<n;i++)        {            printf("%d ",dp[i]);        }        printf("\n");    }}