Parencodings http://acm.pku.edu.cn/JudgeOnline/problem?id=1068
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 8287
Accepted: 4920
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:
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
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Solution
This one is fairly simple, the key is to find the inner connection between this p-sequence and w-sequence. Only need one pass to finish the conversion.
import java.util.Scanner; public class POJ_1068 { public static void main(String[] args) { Scanner in = new Scanner(System.in); int test_num = in.nextInt(); for (int k = 0; k < test_num; k++) { String result = ""; int element_num = in.nextInt(); int leftCounter[] = new int[element_num]; int current = -1, prev = -1; for (int i = 0; i < element_num; i++) { if (prev == -1) { current = in.nextInt(); prev = current; leftCounter[i] = current; } else { prev = current; current = in.nextInt(); leftCounter[i] = current - prev; } } for (int i = 0; i < element_num; i++) { if (leftCounter[i] != 0) { result += "1 "; leftCounter[i]--; } else { for (int j = i - 1; j != -1; j--) { if (leftCounter[j] != 0) { result += (1 + i - j) + " "; leftCounter[j]--; break; } } } } System.out.println(result); } } }