Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17819		Accepted: 10715

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

264 5 6 6 6 69 4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 61 1 2 4 5 1 1 3 9

Source

 

 

#include
     
      using namespace std;void print_int_array(int *int_array, int length){
     	for(int i = 0; i < length; ++i){
     		cout<
      
       <<" ";	}	cout<
       
         k)&&(b[i] == 0))--j;		k = j;		result[i] = i - k + 1;		--b[k];		while((b[k] == 0)&& (k >= 0))--k;	}	return result;}int main(){
     	int m;	cin>>m;	for(int s = 0; s < m; ++s){
     		int n;		int k = 0;		cin>>n;		int *a = new int[n];		for(int i = 0; i < n; ++i){
     			cin>>a[i];		}		print_int_array(calculate(a, n), n);	}	return 0;}