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[];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        p[]=;
        for(int i=;i<=n;i++){
            scanf("%d",&p[i]);
        }
        for(int i=;i<=n;i++){
            for(int k=;k<=i;k++){
                if(p[i]-p[i-k]>=k){
                    printf("%d ",k);
                    break;
                }
            }
        }
        printf("\n");
    }
    return ;
}