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[30] , w[30] ;
int n ;
char s[200] ;

void convert1 ()
{
    int k = 0 ;
    for (int i = 0 ; i < p[0] ; i++)
        s[k++] = '(' ;
    s[k++] = ')' ;
    for (int i = 1 ; i < n ; i++) {
        for (int j = 0 ; j < p[i] - p[i - 1]; j++) {
            s[k++] = '(' ;
        }
        s[k++] = ')' ;
    }
    s[k] = '\0' ;
   // puts (s) ;
}
void convert2 ()
{
    int l , r ;
    int k = 0 ;
    for (int i = 0 ; s[i] != '\0' ; i++) {
        if (s[i] == ')') {
            l = 0 ;
            r = 1 ;
            for (int j = i - 1 ; j >= 0 ; j--) {
                if (s[j] == ')' )
                    r++ ;
                else
                    l++ ;
                if (l == r)
                    break ;
            }
            w[k++] = r;
        }
    }
    for (int i = 0 ; i < n ; i++) {
        printf ("%d" , w[i]) ;
        if (i != n - 1)
            printf (" ") ;
    }
    puts ("") ;
}
int main ()
{
   // freopen ("a.txt" , "r" , stdin) ;
    int T ;
    scanf ("%d" , &T) ;
    while (T--) {
        scanf ("%d" , &n) ;
        for (int i = 0 ; i < n ; i++)
            scanf ("%d" , &p[i]) ;
        convert1 () ;
        convert2 () ;
    }
}