Parencodings
Time Limit: 1000MS 

Memory Limit: 10000K 
Total Submissions: 20999 

Accepted: 12585 
Description
Let S = S1 s2...s2n is a wellformed string of parentheses. S can is encoded in different ways:
QBy an integer sequence p = p1 p2...pn where pi are the number of left parentheses before the ith right parenthesis in S (P sequence).
QBy 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. (Wsequence).
Following is an example of the above encodings:
S ((((() () () ()))
Psequence 4 5 6666
Wsequence 1 1 1456
Write a program to convert psequence of a wellformed string to the wsequence of the same string.
Input
The first line of the input contains a single integer t (1 <= t <=), the number of test cases, followed by the Put data for each test case. The first line of all test case was an integer n (1 <= n <=), and the second line is the psequence of a wellfor Med string. It contains n positive integers, separated with blanks, representing the psequence.
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 wsequence of the string corresponding to its Given psequence.
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
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <cmath > #include <cstdlib> #include <algorithm>using namespace Std;int n,a[21],sum[21];int main () { int tt; scanf ("%d", &tt); while (tt) { scanf ("%d", &n); for (int i=1;i<=n;i++) { scanf ("%d", &a[i]); } for (int i=1;i<=n;i++) sum[i]=a[i]a[i1]; for (int i=1;i<=n;i++) { int j=i; while (J>1&&!sum[j]) j; sum[j]; printf ("%d%c", ij+1,i==n? \ n ': '); } } return 0;}
(analog) POJ 1068