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Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19283		Accepted: 11629

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

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AC代碼例如以下：

#include<iostream>
using namespace std;
int main()
{
    int t,n;
    int i,j;
    int b[30],c[31];
    char a[30];
    cin>>t;
    while(t--)
    {
        cin>>n;
        for(i=1,b[0]=0;i<=n;i++)
            {
                cin>>b[i];
                c[i]=b[i]-b[i-1];
            }
        int tt=0;
        for(i=1;i<=n;i++)//生成匹配括號
        {
            for(j=1;j<=c[i];j++)
                a[tt++]=‘(‘;
            a[tt++]=‘)‘;
        }
        for(i=0;i<tt;i++)
            if(a[i]==‘)‘)
            b[i]=1;
            else b[i]=-1;
        int sum,ans;
        for(i=0;i<tt;i++)
        {
            if(b[i]==1)
            {
               ans=0;sum=0;
               for(j=i;j>=0;j--)
               {
                   sum+=b[j];
                   ans++;
                   if(sum==0)
                    {
                        if(i!=tt-1)
                        cout<<ans/2<<" ";
                        else cout<<ans/2;
                        break;
                    }
               }
            }

        }
        cout<<endl;
    }
    return 0;
}

POJ 1068 Parencodings