Parencodings

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



这个题理解了题意就不难了

输入2 6 不解释

4 5 6 6 6 6

可以这样理解

（（（（）（）（））））

第一个右括号左边有4个左括号

第二个右括号左边有5个左括号

以此类推；

所以根据所给出的数字串；很容易还原括号；

输出时：（（（（）（）（））））



一个完整的（）输出1；共3个

（（）（）（））这个输出4一个（）有3个（），3+1；



一句话就是

比如(((()()())))，
输入：每个右括号之前的左括号数序列为P=4 5 6 6 6 6，；

输出：每个右括号所在的括号内包含的括号数为W=1 1 1 4 5 6.


可以利用 栈和队列

#include<iostream>
#include<queue>
#include<stack>
#include<cstdio>
using namespace std;
int main()
{
    queue<char>p,q;
    stack<char>Q;
    int a,b,c,m,n;
    cin>>n;
    while(n--)
    {
        cin>>m;
        b=0;
        while(m--)
        {
            cin>>a;
            c=a-b;//制定压入的左括号的数目
            b=a;
            while(c--)
                p.push('(');
            p.push(')');//每次压入左括号之后压入一个右括号
        }
        while(!p.empty())//将队列中的括号一个个取出
        {
            if(p.front()=='(')
                Q.push(p.front());//如果是左括号直接压入栈中
            
            else if(p.front()==')')//如果是右括号，则进行判断
            {
                if(Q.top()=='(')//如果栈顶元素是左括号
                {
                    Q.pop();//出栈
                    Q.push(1);//压入 1
                    q.push(1);//这是为了把数字储存下来，方便以后输出答案
                }
                else//如果栈顶元素不是左括号
                {
                    int sum=1;
                    while(Q.top()!='(')//持续出栈  直到左括号
                    {
                        sum+=Q.top();//统计一共包含多少左括号
                        Q.pop();
                    }
                    Q.pop();//左括号出栈
                    q.push(sum);//这是为了把数字储存下来，方便以后输出答案
                    Q.push(sum);//压入括号个数
                }
            }
            p.pop();
        }
        printf("%d",q.front());
        q.pop();
        while(!q.empty())
        {
            printf(" %d",q.front());
            q.pop();
        }
        printf("\n");
    }
    return 0;
}