1.Link:

http://poj.org/problem?id=1068

http://bailian.openjudge.cn/practice/1068

2.Content:

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20077		Accepted: 12122

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

264 5 6 6 6 69 4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 61 1 2 4 5 1 1 3 9

Source

Tehran 2001

3.Method:

（1）先将P-sequence转成S序列，方法为：利用vector保存，置入）前，放入当前数值减前一数值的（

（2）将S序列转为W序列，方法为：利用stack，每次遇到（则置入stack，其中-1代表“（”；遇到count = 1，直到遇到第一个（前，将stack的值累加到count，并置出

4.Code:

 1 #include <iostream> 2 #include <vector> 3 #include <stack> 4  5 using namespace std; 6  7 int main() 8 { 9     //freopen("D://input.txt","r",stdin);10 11     int i,j;12 13     int t;14     cin >> t;15     while(t--)16     {17         int n;18         cin >> n;19 20         int *arr_p = new int[n];21 22         for(i = 0; i < n; ++i) cin >> arr_p[i];23 24         //for(i = 0; i < n; ++i) cout << arr_p[i] << endl;25 26 27         vector<char> v_sym;28 29         int pre_p = 0;30         for(i = 0; i < n; ++i)31         {32             for(j = pre_p; j < arr_p[i]; ++j) v_sym.push_back(‘(‘);33             v_sym.push_back(‘)‘);34             pre_p = arr_p[i];35         }36 37         vector<char>::size_type sym_i;38 39         //for(sym_i = 0; sym_i != v_sym.size(); ++sym_i) cout << v_sym[sym_i];40         //cout << endl;41 42         stack<int> s_sym;// -1 (, -2 )43         for(sym_i = 0; sym_i != v_sym.size(); ++sym_i)44         {45             if(‘(‘ == v_sym[sym_i]) s_sym.push(-1);46             else47             {48                 int count = 1;49                 while(s_sym.top() != -1)50                 {51                     count += s_sym.top();52                     s_sym.pop();53                 }54                 s_sym.pop();55                 cout << count << " ";56                 s_sym.push(count);57             }58         }59         cout << endl;60 61 62 63         delete [] arr_p;64 65 66     }67 68     return 0;69 }

5.Reference:

Poj OpenJudge 1068 Parencodings