POJ 1068 Parencodings

Parencodings
Time Limit: 1000 MS Memory Limit: 10000 K
Total Submissions: 15566 Accepted: 9275
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. (W-sequence ).

Following is an example of the above encodings:
S (((()()())))

P-sequence 4 5 6666

W-sequence 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 shoshould 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
9
4 6 6 6 6 8 9 9 9
Sample Output

1 1 1 4 5 6
1 1 2 4 5 1 3 9
Source

Tehr 2001
Site: Stack
Just read the question.
[Cpp]
# Include <iostream>
# Include <cstring>
# Include <stdio. h>
# Include <math. h>
Using namespace std;
Int a [100], top, res [100], top2;
Class str
{
Public:
Int pos;
Char c;
} S1 [100];
Class statck
{
Public:
Int pos;
Char c;
} S2 [1, 100];
Int main ()
{
Int I, j, n, m, s, t, pos1, pos2;
Cin> t;
While (t --)
{
Cin> n;
For (I = 0; I <= n-1; I ++)
{
Cin> a [I];
}
Top = 0;
For (I = 0; I <= n-1; I ++)
{
If (! I)
{
For (j = 1; j <= a [I]; j ++)
{
S1 [top]. c = '(';
S1 [top]. pos = top;
Top ++;
}
S1 [top]. c = ')';
S1 [top]. pos = top;
Top ++;
} Else
{
For (j = 1; j <= (a [I]-a [I-1]); j ++)
{
S1 [top]. c = '(';
S1 [top]. pos = top;
Top ++;
}
S1 [top]. c = ')';
S1 [top]. pos = top;
Top ++;
}
}
N = top;
Top = 0;
Top2 = 0;
Memset (res, 0, sizeof (res ));
For (I = 0; I <= n-1; I ++)
{
If (s1 [I]. c = '(')
{
S2 [top]. c = '(';
S2 [top]. pos = s1 [I]. pos;
Top ++;
} Else
{
Pos1 = s2 [top-1]. pos;
Top-= 1;
Pos2 = s1 [I]. pos;
For (j = pos1; j <= pos2; j ++)
{
If (s1 [j]. c = ')')
{
Res [top2] + = 1;
}
}
Top2 ++;
}
}
For (I = 0; I <= n/2-1; I ++)
{
If (! I)
{
Cout <res [I];
} Else
{
Cout <"<res [I];
}
}
Cout <endl;
}
Return 0;
}