poj 1068 Parencodings

Parencodings

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

Source

Tehran 2001

 1 //poj  1068
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <queue>
 5 #include <stack>
 6 #include <iostream>
 7 using namespace std;
 8 stack<int> s;
 9 int main(){
10     //freopen("D:\\INPUT.txt","r",stdin);
11     int t,n;
12     cin>>t;
13     while(t--){
14         while(!s.empty()){
15             s.pop();
16         }
17         cin>>n;
18         int i,j,k,num,pre=0;
19         for(i=0;i<n;i++){
20             if(i){
21                cout<<" ";
22             }
23             scanf("%d",&num);
24             //cout<<num<<endl;
25             j=num-pre;//相邻两个右括号之间的左括号数目-1
26             pre=num;//记录前一个右括号前面哪有几个左括号
27             if(j==1){//多一个左括号
28                 cout<<1;
29                 if(!s.empty()){
30                         s.top()++;
31                 }
32             }
33             else{
34                 if(!j){//相等
35                     int next=++s.top();
36                     cout<<s.top();
37                     s.pop();
38                     if(!s.empty()){
39                         s.top()+=next;
40                     }
41                 }
42                 else{//隔了不止一个左括号
43                     cout<<1;
44                     j--;
45                     while(j--){
46                         s.push(0);
47                     }
48                     s.top()++;
49                 }
50             }
51         }
52         cout<<endl;
53         //cout<<"aa"<<endl;
54     }
55     return 0;
56 }