Codeforces Round #162 (Div. 2) C

C. Escape from Stones

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones
will fall and Liss will escape from the stones. The stones are numbered from 1 to n in order.

The stones always fall to the center of Liss's interval. When Liss occupies the interval [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].

You are given a string 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 stones falls.

Input

The input consists of only one line. The only line contains the string s (1 ≤ |s| ≤ 106).
Each character in s will be either "l" or "r".

Output

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.

题目大意:就是在[0,1]区间掉石头的时候,Squirrel Liss就在逃,石头都是往区间的中间砸,然后Squirrel Liss就左右左右的逃.求把石头从左到右的石头编码输出来

思路:一开始用模拟真心NC,精度完全过不了但就是眼睁睁的跳进坑里..后来看大大的解题报告晓得这么一个规律:

当选择左边的时候,那么之后掉下来的石头一定是在之前的石头的左边

当选择右边的时候,那么之后掉下来的石头一定是在之前的石头的右边(因为人走哪边石头就砸哪边)

AC program:

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int position[1000010];
char s[1000010];
int main()
{
scanf("%s",s);
int len=strlen(s);
int left=0,right=len-1;
for(int i=0;i<len;i++)
{
if(s[i]=='l')
position[right--]=i+1;
else
position[left++]=i+1;
}
for(int i=0;i<len;i++)
{
printf("%d\n",position[i]);
}
//system("pause");
return 0;}