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

 1 #include<iostream>

 2 #include<queue>

 3 #include<stack>

 4 #include<cstdio>

 5 using namespace std;

 6 int main()

 7 {

 8     queue<char>p,q;

 9     stack<char>Q;

10     int a,b,c,m,n;

11     cin>>n;

12     while(n--)

13     {

14         cin>>m;

15         b=0;

16         while(m--)

17         {

18             cin>>a;

19             c=a-b;//制定压入的左括号的数目

20             b=a;

21             while(c--)

22                 p.push('(');

23             p.push(')');//每次压入左括号之后压入一个右括号

24         }

25         while(!p.empty())//将队列中的括号一个个取出

26         {

27             if(p.front()=='(')

28                 Q.push(p.front());//如果是左括号直接压入栈中

29             

30             else if(p.front()==')')//如果是右括号，则进行判断

31             {

32                 if(Q.top()=='(')//如果栈顶元素是左括号

33                 {

34                     Q.pop();//出栈

35                     Q.push(1);//压入 1

36                     q.push(1);//这是为了把数字储存下来，方便以后输出答案

37                 }

38                 else//如果栈顶元素不是左括号

39                 {

40                     int sum=1;

41                     while(Q.top()!='(')//持续出栈  直到左括号

42                     {

43                         sum+=Q.top();//统计一共包含多少左括号

44                         Q.pop();

45                     }

46                     Q.pop();//左括号出栈

47                     q.push(sum);//这是为了把数字储存下来，方便以后输出答案

48                     Q.push(sum);//压入括号个数

49                 }

50             }

51             p.pop();

52         }

53         printf("%d",q.front());

54         q.pop();

55         while(!q.empty())

56         {

57             printf(" %d",q.front());

58             q.pop();

59         }

60         printf("\n");

61     }

62     return 0;

63 }

View Code