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  • Codeforces 264A. Escape from Stones

    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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  • 原文地址:https://www.cnblogs.com/ToTOrz/p/7260810.html
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