Codeforces Round #162 (Div. 1) A. Escape from Stones

A. 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.

1（）题目大意 ， 

就是说 一个 一个人 站在一个 0 - 1的线段上， 最开始站在中点处，然后会有炸弹不断的向下落到他当前的位置，他会向左/右跑每次移动的距离缩小到上一次的1/2，（第一次1/4，第二次1/8........）

输出 ，在这条线段上的每个点处（被炸弹轰过的地方）的顺序是什么 。。。具体看例子把..只能说到这个地步了

（2）思路，这道题我想的是只考虑当前点，那么如果向右，那么这个点就是当前的最后一点，因为每次移动的距离缩小一半，也就是说她之后无论怎么向左移动，都无法到达当前点，同理，向左移动的话是一样的

 1 #include <algorithm>
 2 #include <stack>
 3 #include <istream>
 4 #include <stdio.h>
 5 #include <map>
 6 #include <math.h>
 7 #include <vector>
 8 #include <iostream>
 9 #include <queue>
10 #include <string.h>
11 #include <set>
12 #include <cstdio>
13 #define FR(i,n) for(int i=0;i<n;i++)
14 #define MAX 2005
15 #define mkp pair <int,int>
16 using namespace std;
17 const int maxn = 1e6+5;
18 typedef long long ll;
19 const int  inf = 0x3fffff;
20 void read(int  &x) {
21     char ch; bool flag = 0;
22     for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
23     for (x = 0; isdigit(ch); x = (x << 1) + (x << 3) + ch - 48, ch = getchar());
24     x *= 1 - 2 * flag;
25 }
26 
27 int ans[maxn];
28 char s1[maxn];
29 int main() {
30     cin>>s1;
31 
32     int len = strlen(s1);
33     int st = 0, last = len;
34     for(int i=0;i<len;i++){
35         if(s1[i]=='r'){
36             ans[st++]=i+1;
37         }
38         else {
39             ans[--last]=i+1;
40         }
41     }
42     for(int i=0;i<len;i++){
43         printf("%d\n",ans[i]);
44     }
45     return 0;
46 }