proj1068

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

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<iostream>

using namespace std;

void print_int_array(int *int_array, int length){
	for(int i = 0; i < length; ++i){
		cout<<int_array[i]<<" ";
	}
	cout<<endl;
}

int *calculate(int *a, int n){
	int k = 0;
	int *b = new int[n];
	int *result = new int[n];
	b[0] = a[0];
	for(int i = 1; i < n; ++i){
		b[i] = a[i] - a[i - 1];
	}
	for(int i = 0; i < n; ++i){
		int j = i;
		while((j > 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;
}