Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 25090		Accepted: 14793

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

感觉比较难想，题也是读不懂，看了网上人家的翻译，第一行数据是右括号的左括号数，第二行是右括号匹配的括号中包括几对括号，包括它本身。

#include<iostream>
using namespace std;
int main()
{
	int a[30],b[30],c[30];
	int t,n;
	cin >> t;
	while(t--)
	{
		cin >> n;
		for(int i = 0;i < n;i++)
			cin >> a[i];
		b[0] = a[0];
		for(int i = 1;i < n;i++)
		{
			b[i] = a[i] - a[i-1];
		}
		for(int i = 1;i <= n;i++)
		{
			int j;
			for(j = i - 1;j >= 0;j--)
			{
				if(b[j] > 0)
				{
					b[j]--;
					break;
				}
			}
			c[i] = i - j;
		}
		for(int i = 1;i <= n;i++)
		{
			if(i == n)
				cout << c[i] << endl;
			else
				cout << c[i] << ' ';
		}
	}
	return 0;
}