[JAVA][ZOJ 1016][Parencodings]

Let S = s1 s2 ... s2n be a well-formed string of parentheses. S can be encoded in two different ways:

• 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).
• 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 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

[java] view
plaincopyprint?

1. <p>
2. </p>import java.io.BufferedInputStream;
3. import java.util.Scanner;
4. public class Main {
5. public static void main(String[] args) throws InterruptedException {
6. Scanner sc = new Scanner(System.in);
7. int n = sc.nextInt();
8. for (int i = 0; i < n; i++) {
9. int len = sc.nextInt();
10. int P[] = new int[len];// P shows how many left parentheses have been seen since each right parentheses
11. int W[] = new int[len];// W shows how many left parentheses have been seen within a pair of parentheses
12. // EG:<><>,W=11
13. // EG:<<>>,W=12
14. // EG:<<<><>>>,W=1134
15. char str[] = new char[len * 2];
16. int posInP = 0;
17. for (int posInStr = 0; posInStr < len; posInStr++) {
18. P[posInStr] = sc.nextInt();
19. if (posInStr == 0) {// the first P-code
20. for (; posInP < P[posInStr]; posInP++) {
21. str[posInP] = ‘<‘;
22. }
23. str[posInP++] = ‘>‘;
24. } else {
25. if (P[posInStr] > P[posInStr - 1]) {
26. for (int k = 0; k < (P[posInStr] - P[posInStr - 1]); k++) {
27. str[posInP++] = ‘<‘;
28. }
29. str[posInP++] = ‘>‘;
30. } else {// if the next num is the same of the last
31. str[posInP++] = ‘>‘;
32. }
33. }
34. }
35. // check for W array
36. int posInW = 0;
37. for (int j = 0; j < len * 2; j++) {
38. if (str[j] == ‘<‘) {
39. continue;
40. }
41. int RightParentheses = 1;
42. for (int j2 = j - 1; j2 >= 0; j2--) {
43. if (str[j2] == ‘>‘) {
44. RightParentheses++;
45. } else {
46. RightParentheses--;
47. W[posInW]++;
48. }
49. if (RightParentheses == 0) {
50. posInW++;
51. break;
52. }
53. }
54. }
55. int j;
56. for (j = 0; j < len - 1; j++) {
57. System.out.print(W[j] + " ");
58. }
59. System.out.print(W[j]);
60. System.out.println();
61. }
62. }
63. }

PE了好多次，在Eclipse中复制2个testcase，刚复制完就出现第一个的结果，然后按回车就出现另一个结果，本来以为是这个问题，改为把所有的结果收集然后集中输出后又显示RE，最后发现居然是输出的问题，最后一个数字后面不输出空格，这细节真心不是坑爹嘛。。。。