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.
补一补昨晚的训练题。挺长时间没有写博客了，心里满满的负罪感。这个题目的意思不难理解，就是一个简单的二分。我一开始就是按照题目的要求，一点一点的分下去，样例过了，但是提交一发就wa了。我想了想，数据量是1e6╮(╯﹏╰）╭就一个数，分1e6次也会接近零，试了试long double，还是不行。就想新的方法了。还有要注意的就是cin和cout超时，在这儿又wa了一发。
代码如下：
#include<bits/stdc++.h>
using namespace std;
const int maxx=1e6+10;
char a[maxx];
struct node{
long double pos;
int ant;
}p[maxx];
int cmp(const node &a,const node &b)
{
return a.pos<b.pos;
}
int main()
{
while(scanf("%s",a)!=EOF)
{
int len=strlen(a);
int maxx=len;
int minn=1;
for(int i=0;i<len;i++)
{
if(a[i]=='l')
{
p[i].pos=maxx--;
p[i].ant=i+1;
}
else
{
p[i].pos=minn++;
p[i].ant=i+1;
}
}
sort(p,p+len,cmp);
for(int i=0;i<len;i++)
printf("%d\n",p[i].ant);
//cout<<p[i].ant<<endl;
}
}
努力加油a啊，(o)/~