Parencodings(imitate)

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20679		Accepted: 12436

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

 #include<stdio.h>
 #include<string.h>
 #include<iostream>
 using namespace std;
 int p[] , w[] ;
 int n ;
 char s[] ;

 void convert1 ()
 {
     int k =  ;
     for (int i =  ; i < p[] ; i++)
         s[k++] = '(' ;
     s[k++] = ')' ;
     for (int i =  ; i < n ; i++) {
         for (int j =  ; j < p[i] - p[i - ]; j++) {
             s[k++] = '(' ;
         }
         s[k++] = ')' ;
     }
     s[k] = '\0' ;
    // puts (s) ;
 }
 void convert2 ()
 {
     int l , r ;
     int k =  ;
     for (int i =  ; s[i] != '\0' ; i++) {
         if (s[i] == ')') {
             l =  ;
             r =  ;
             for (int j = i -  ; j >=  ; j--) {
                 if (s[j] == ')' )
                     r++ ;
                 else
                     l++ ;
                 if (l == r)
                     break ;
             }
             w[k++] = r;
         }
     }
     for (int i =  ; i < n ; i++) {
         printf ("%d" , w[i]) ;
         if (i != n - )
             printf (" ") ;
     }
     puts ("") ;
 }
 int main ()
 {
    // freopen ("a.txt" , "r" , stdin) ;
     int T ;
     scanf ("%d" , &T) ;
     while (T--) {
         scanf ("%d" , &n) ;
         for (int i =  ; i < n ; i++)
             scanf ("%d" , &p[i]) ;
         convert1 () ;
         convert2 () ;
     }
 }