POJ1068

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 25217 Accepted: 14877

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 6666W-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

使用stack，将p序列还原成括号序列

将左括号记为1，压入栈，扫到右括号，弹出栈顶元素，将新的栈顶元素加上已扫到的右括号的数目，

弹出数字序列即为w序列

#include <iostream>#include <string>#include <stack>using namespace std;int main(){    int n;    cin >> n;    for (int i=0; i<n; i++)    {        int m;        cin >> m;        string str;        int leftpa = 0;        for (int j=0; j<m; j++)        {            int p;            cin >> p;            for (int k=0; k<p-leftpa; k++) str += '(';            str += ')';            leftpa = p;        }        stack<int> s;        for (string::iterator it=str.begin(); it!=str.end(); it++)        {            if (*it=='(')                s.push(1);            else            {                int p = s.top(); s.pop();                cout << p << " ";                if (!s.empty()) s.top() += p;            }        }        cout << endl;    }    return 0;}