Parencodings(POJ1068 模拟)

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 27179 Accepted: 15973

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

#include <iostream>#include <stack>using namespace std;stack<int> s;int main(){int t, n, k, m;int i, j;cin >> t;bool flag[1000] = { false };//flag还原左右括号位置for (i = 0; i<t; i++){cin >> n;memset(flag, 0, 1000);k = 0;for (j = 0; j<n; j++){cin >> m;flag[k + m] = true;//右括号位置设为truek++;}for (j = 0; j<k + m - 1; j++)//j记录位置{if (!flag[j])//左括号位置入栈{s.push(j);}else{cout << (j - s.top() + 1) / 2 << " ";//匹配的左右括号之间的括号对数除以2s.pop();//已匹配的括号出栈，让s.top()为左边左括号的位置}}cout << (j - s.top() + 1) / 2 << endl;s.pop();}return 0;}

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