Parencodings(imitate)

Parencodings

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

Source

Tehran 2001

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<iostream>
 4 using namespace std;
 5 int p[30] , w[30] ;
 6 int n ;
 7 char s[200] ;
 8
 9 void convert1 ()
10 {
11     int k = 0 ;
12     for (int i = 0 ; i < p[0] ; i++)
13         s[k++] = ‘(‘ ;
14     s[k++] = ‘)‘ ;
15     for (int i = 1 ; i < n ; i++) {
16         for (int j = 0 ; j < p[i] - p[i - 1]; j++) {
17             s[k++] = ‘(‘ ;
18         }
19         s[k++] = ‘)‘ ;
20     }
21     s[k] = ‘\0‘ ;
22    // puts (s) ;
23 }
24 void convert2 ()
25 {
26     int l , r ;
27     int k = 0 ;
28     for (int i = 0 ; s[i] != ‘\0‘ ; i++) {
29         if (s[i] == ‘)‘) {
30             l = 0 ;
31             r = 1 ;
32             for (int j = i - 1 ; j >= 0 ; j--) {
33                 if (s[j] == ‘)‘ )
34                     r++ ;
35                 else
36                     l++ ;
37                 if (l == r)
38                     break ;
39             }
40             w[k++] = r;
41         }
42     }
43     for (int i = 0 ; i < n ; i++) {
44         printf ("%d" , w[i]) ;
45         if (i != n - 1)
46             printf (" ") ;
47     }
48     puts ("") ;
49 }
50 int main ()
51 {
52    // freopen ("a.txt" , "r" , stdin) ;
53     int T ;
54     scanf ("%d" , &T) ;
55     while (T--) {
56         scanf ("%d" , &n) ;
57         for (int i = 0 ; i < n ; i++)
58             scanf ("%d" , &p[i]) ;
59         convert1 () ;
60         convert2 () ;
61     }
62 }