Parencodings

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 25090 Accepted: 14793

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

感觉比较难想，题也是读不懂，看了网上人家的翻译，第一行数据是右括号的左括号数，第二行是右括号匹配的括号中包括几对括号，包括它本身。

#include<iostream>using namespace std;int main(){int a[30],b[30],c[30];int t,n;cin >> t;while(t--){cin >> n;for(int i = 0;i < n;i++)cin >> a[i];b[0] = a[0];for(int i = 1;i < n;i++){b[i] = a[i] - a[i-1];}for(int i = 1;i <= n;i++){int j;for(j = i - 1;j >= 0;j--){if(b[j] > 0){b[j]--;break;}}c[i] = i - j;}for(int i = 1;i <= n;i++){if(i == n)cout << c[i] << endl;elsecout << c[i] << ' ';}}return 0;}