Codeforces 264A. Escape from Stones

Posted 2020-09-30 TOTOTOTOTZZZZZ

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 swill 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.

最初用vector做的，一直超，后来换了数组发现还是超，才想到是cin,cout的问题，加了一句sync果然好多了

#include <cstdio>
#include <cctype>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
using namespace std;
typedef long long LL;


int total = 0;
int lv[1000005] = {0};
string s;
int main() {
  // freopen("test.in","r",stdin);
  ios::sync_with_stdio(false);
  cin >> s;
  int len = s.length();
  for (int i=0;i<len;i++){
    int now = i+1;
    if (s[i] == ‘l‘){
      total ++;
      lv[total] = now;
    }
    else {
      cout << now << endl;
    }
  }
  for (int i=total;i>=1;i--){
    cout << lv[i] << endl;
  }

  return 0;
}

View Code

 

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