CF 264A(向内的双向队列)

C. Escape from Stones

time limit per test

memory limit per test

input

output

[0, 1]. Next, n stones will fall and Liss will escape from the stones. The stones are numbered from 1 to n

[k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the right, her new interval will be [k, k + d].

s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n

Input

s (1 ≤ |s| ≤ 10⁶). Each character in s will be either "l" or "r".

Output

n lines — on the i-th line you should print the i-th stone's number from the left.

Sample test(s)

input

llrlr

output

3 5 4 2 1

input

rrlll

output

1 2 5 4 3

input

lrlrr

output

2 4 5 3 1

Note

In the first example, the positions of stones 1, 2, 3, 4, 5 will be 

, respectively. So you should print the sequence: 3, 5, 4, 2, 1.

不能用模拟double+除法，会爆精度啊！！（long double 也不行）

其实只要根据性质，在序列前后添加即可。

靠，人生中的处女Hack，竟然是被Hack…(受？）

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<functional>
#include<algorithm>
#include<cctype>
using namespace std;
#define MAXN (1000000+10)
//pair<double,int> a[MAXN];
char s[MAXN];
int n,a[MAXN];
int main()
{
  scanf("%s",&s);
  n=strlen(s);
  int l=1,r=n;
  for (int i=0;i<n;i++)
  {
    if (s[i]=='l') a[r--]=i+1;
    else a[l++]=i+1;        
  }
  
  
  
  for (int i=1;i<=n;i++) cout<<a[i]<<endl;
  
  
  
}