CodeForces - 264A（思維題）

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.

Examples
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.

• 題目大意：
  有一個[0,1]的區間，然後每次砸石頭砸到區間的中間，然後一個人如果選擇往右跳那區間就更新為 [0.5,1] 如果往左跳區間就更新為[0,0.5]。然後下一次再砸到中間……依次往下。把石頭都標上號，然後輸出的是從左到右石頭的 標號數。
• 解題思路：
  ans[i]裡存的是第i個位置碰到的石頭編號
  假如，第一次你往左跳的話，那麼ans[n]=1;因為不可能有其他的石頭會在這個石頭的右邊了因為區間更新為[0,0.5]了，所以第n個石頭一定是 1號石頭。同樣的如果第一次往有跳，那麼ans[1]=1 因為不可能有石頭會在他的左邊了，因為區間更新為[0.5,1]了。然後每一個石頭都這麼考慮。。類似於一個雙向的佇列。
• AC程式碼

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=1e6+10;
int l,r;
char a[maxn];
int ans[maxn];
int main()
{
	scanf("%s",a);
	int len=strlen(a);
	l=1;r=len;
	for(int i=0;i<len;i++)
	{
		if(a[i]=='l')
			ans[r--]=i+1;
		else
			ans[l++]=i+1;
	}
	for(int i=1;i<=len;i++)
		//cout<<ans[i]<<endl;
		printf("%d\n",ans[i]);
	return 0;
}