poj 1068Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20019		Accepted: 12080

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

Source

Tehran 2001

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>

using namespace std;
int a[21];
int ans[21];
int main()
{
    int Case,n,tmp;
    string s;
    cin>>Case;
    while(Case--)
    {
        cin>>n;
        tmp=0;
        for(int i=0; i<n; i++)
        {
            cin>>a[i];
            while(tmp<a[i])
            {
                tmp++;
                s+='(';
            }
            s+=')';
            int left=1,right=1;
            for(int j=s.length()-2;j>=0&&right;j--)
                {
                    if(s[j]==')')
                    {
                        left++;
                        right++;
                    }
                     else right--;
                }
                ans[i]=left;
        }
        //cout<<s<<endl;
        for(int i=0;i<n;i++)
        {
            cout<<ans[i]<<" ";
            if(i==n-1)
                cout<<endl;
        }
     //cout<<sizeof(s)<<endl;
    }
    return 0;
}