Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19068		Accepted: 11503

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 简单模拟题，给出若干左右括号，能写出两种数列，其一:在第i个右括号前有多少个左括号pi就是几。 其二：找距离第i个右括号最近的左括号（没有配过对的）。 此题给出其一队列，求其二，模拟即可，不多说。

1 /*====================================================================== 2  *           Author :   kevin 3  *         Filename :   Parencodings.cpp 4  *       Creat time :   2014-05-23 13:41 5  *      Description : 6 ========================================================================*/ 7 #include 
      
        8 #include 
       
         9 #include 
        
         10 #include 
         
          11 #include 
          
           12 #include 
           
            13 #define clr(a,b) memset(a,b,sizeof(a))14 #define M 5015 using namespace std;16 char str[M];17 int vis[M];18 int main(int argc,char *argv[])19 {20 int t,n;21 scanf("%d",&t);22 while(t--){23 scanf("%d",&n);24 clr(str,0);25 clr(vis,0);26 int a,cc = 0,cnt = 0;27 for(int i = 0; i < n; i++){28 scanf("%d",&a);29 for(int j = cc; j < a; j++){30 str[cnt++] = '(';31 }32 str[cnt++] = ')';33 cc = a;34 }35 int kong = 0;36 for(int i = 0; i < 2*n; i++){37 if(str[i] == ')'){38 int steps = 1;39 for(int j = i-1; j >= 0; j--){40 if(str[j] == '(' && !vis[j]){41 if(kong != n-1){42 printf("%d ",steps);43 kong++;44 }45 else{46 printf("%d\n",steps);47 }48 vis[j] = 1;49 break;50 }51 if(str[j] == '(' && vis[j]){52 steps++;53 }54 }55 }56 }57 }58 return 0;59 }