# (analog) POJ 1068

Source: Internet
Author: User

Parencodings
 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 20999 Accepted: 12585

Description

Let S = S1 s2...s2n is a well-formed 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. (W-sequence).

Following is an example of the above encodings:
`S ((((() () () ()))P-sequence    4 5 6666W-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 <=), 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 p-sequence of a well-for Med 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`
`#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[i-1];        for (int i=1;i<=n;i++)        {            int j=i;            while (J>1&&!sum[j])                j--;            sum[j]--;            printf ("%d%c", i-j+1,i==n? \ n ': ');        }    }    return 0;}`

(analog) POJ 1068

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