POJ 1068 Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19409		Accepted: 11718

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

#include<iostream>
#include<cstring>
using namespace std;

int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		int n;
		int i,j;
		int a[22],b[45],c[22];
		memset(b,0,sizeof(b));
		cin>>n;
		int d,k=1;
		a[0]=0;

		for(i=1;i<=n;i++)
		{
			cin>>a[i];
			d=a[i]-a[i-1];
			b[d+a[i-1]+k++]=1;
		}

		int ans,p=0;
		for(i=1;i<=2*n;i++)
		{
			if(b[i]==1)
			{
				b[i]=3;
				for(j=i-1;j>0;j--)
				{			
					ans=0;
					if(b[j]==0)  
					{
						b[j]=2;
						for(k=j;k<=i;k++)
							if(b[k]==3)  ans+=1;
						c[++p]=ans;
						break;
					}
				}
			}
		}

		for(i=1;i<p;i++)
			cout<<c[i]<<" ";
		cout<<c[p]<<endl;	
	}

	return 0;
}