poj1068
Parencodings
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 18785 | Accepted: 11320 |
Description
Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways:
q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
q By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).
Following is an example of the above encodings:
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
q By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).
Following is an example of the above encodings:
S (((()()())))
P-sequence 4 5 6666
W-sequence 1 1 1456
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
Input
The
first line of the input contains a single integer t (1 <= t <=
10), the number of test cases, followed by the input data for each test
case. The first line of each test case is an integer n (1 <= n <=
20), and the second line is the P-sequence of a well-formed string. It
contains n positive integers, separated with blanks, representing the
P-sequence.
Output
The
output file consists of exactly t lines corresponding to test cases.
For each test case, the output line should contain n integers describing
the W-sequence of the string corresponding to its given P-sequence.
Sample Input
2 6 4 5 6 6 6 6 9 4 6 6 6 6 8 9 9 9
Sample Output
1 1 1 4 5 6 1 1 2 4 5 1 1 3 9
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
/* * File: main.cpp * Author: liaoyu <liaoyu@whu.edu.cn> * * Created on April 1, 2014, 5:34 PM */ #include <iostream> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <cmath> #include <algorithm> #include<map> using namespace std; int p[23]; int main() { int t; scanf("%d",&t); while(t--){ int n; scanf("%d",&n); p[0]=0; for(int i=1;i<=n;i++){ scanf("%d",&p[i]); } for(int i=1;i<=n;i++){ for(int k=1;k<=i;k++){ if(p[i]-p[i-k]>=k){ printf("%d ",k); break; } } } printf("\n"); } return 0; }
posted on 2014-04-06 20:46 PolarisSherlock 阅读(133) 评论(0) 编辑 收藏 举报