poj1068

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18785		Accepted: 11320

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

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

/* 
 * File:   main.cpp
 * Author: liaoyu <liaoyu@whu.edu.cn>
 *
 * Created on April 1, 2014, 5:34 PM
 */

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <cmath>
#include <algorithm>

#include<map>
using namespace std;
int p[23];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        p[0]=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&p[i]);
        }
        for(int i=1;i<=n;i++){
            for(int k=1;k<=i;k++){
                if(p[i]-p[i-k]>=k){
                    printf("%d ",k);
                    break;
                }
            }
        }
        printf("\n");
    }
    return 0;
}