poj 1068 Parencodings

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21128 Accepted: 12662

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

题目的意思是给出一堆括号的表示方法，其实给出每个右括号的左边又多少个左括号，问每个右括号和其对应的左括号之间的括号有多少，包括自己。

记录每个右括号和前一个右括号之间的左括号，对每个右括号统计一下是否包含另一个呦括号，如果包含就讲坐标左移一位，最后直到不包含，包含的右括号数就是所求。

#include <iostream>using namespace std;int p[233];int l[233];int main(){int T,n;cin>>T;while(T--){cin>>n;for(int i=1;i<=n;i++){cin>>p[i];l[i]=p[i]-p[i-1];int k=i;while(l[k]==0) k--;l[k]--;cout<<i-k+1;if(i!=n) cout<<" ";}cout<<endl;}return 0;}