Parencodings

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 13   Accepted Submission(s) : 7

Problem Description

Let S = s1 s2 … s2n be a well-formed string of parentheses. S can be encoded in two different ways: [ul] [li]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).[/li] [li]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).[/li] [/ul] Following is an example of the above encodings: [pre] S (((()()()))) P-sequence 4 5 6666 W-sequence 1 1 1456 [/pre] 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 file 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

2001 Asia Regional Teheran

 

1 #include 
       
         2 #include 
        
          3  4 int main() 5 { 6     int T,num,i,j,sign,sum,K,k,z,tend,alto; 7     scanf("%d",&T); 8     while(T--) 9     {10         int a[1000]={
     0},b[1000],c[1000]={
     0};11         scanf("%d %d",&num,&sign);12         a[sign]+=1;13         b[0]=sign;14         getchar();15         for(i=sign+1,j=1,k=1;j
         
          =0;z--)38             {39                 if(a[z]==1)40                     continue;41                 else42                 {43                     if(a[z]==0)44                     {a[z]='N';c[j]+=1;break;}45                     else46                     {47                         if(a[z]=='N')48                         {49                             c[j]+=1;50                         }51                             continue;52                     }53                 }54             }55         for(j=0;j