ZOJ 1016 Parencodings *^*

Let S = s1 s2 ... s2n be a well-formed string of parentheses. S can be encoded in two different ways:

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).

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 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

代码如下：

#include <iostream>

#include<stack>

using namespace std;

stack<int>s;

int main()

{

int T,n,p,t;

cin >> T;

while(T--)

{

cin >> n;

t = 0;

for(int i=1; i<=n; i++)

{

cin >> p;

for(int j=1; j<=p-t; j++)

s.push(-1);

if(s.top() == -1)

{

i == 1 ? cout << "1" : cout << " 1";

s.pop();

s.push(1);

}

else

{

int i_top = s.top();

s.pop();

while(s.top() != -1)

{

i_top += s.top();

s.pop();

}

cout << " " << i_top + 1;

s.pop();

s.push(i_top + 1);

}

t = p;

}

cout << endl;

while(!s.empty())

s.pop();

}

return 0;

}