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.
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 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; }