Parencodings - poj 1068

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 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 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

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这道题很简单，一遍AC，用flag数组记录左括号的状态，当是1时已经匹配，当是0时未匹配，向前查找第一个未匹配的左括号，记录中间匹配的左括号的个数

 1 #include <iostream>
 2 using namespace std;
 3
 4 int main() {
 5     int num;
 6     int n;
 7     int p[21];
 8     int flag[21];
 9     cin>>num;
10     for(int i=0;i<num;i++){
11         cin>>n;
12         for(int j=0;j<n;j++){
13             cin>>p[j];
14             flag[j]=0;
15         }
16         for(int k=0;k<n;k++){
17             int count=0;
18             for(int m=p[k]-1;m>=0;m--){
19                 if(flag[m]==0){
20                     cout<<count+1<<" ";
21                     flag[m]=1;
22                     break;
23                 }else{
24                     count++;
25                 }
26             }
27         }
28         cout<<endl;
29     }
30     return 0;
31 }