1044: Parencodings

 1044: Parencodings

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	613	302	Standard

Let S = s1 s2 … s2n be a well-formed string of parentheses. S can be encoded in two different ways:

• 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).
• 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 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

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    int a[50],c[50],ans[50];//-1'('  1')'
    int ci;cin>>ci;
    while(ci--)
    {
        int n;cin>>n;
        for(int i=1;i<=n;i++)cin>>c[i];c[0]=0;
        memset(a,0,sizeof(a));
        int cnt=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=c[i]-c[i-1];j++)
            {
                a[cnt++]=-1;
            }
            a[cnt++]=1;
        }
       // for(int i=1;i<cnt;i++) cout<<a[i]<<"..";cout<<endl;
        int l=0;
        for(int i=1;i<cnt;i++)
        {
            if(a[i]==1)
            {
                a[i]=2;
                int num=0;
                for(int j=i-1;j>=1;j--)
                {
                    if(a[j]==-2) num++;
                    if(a[j]==-1) {a[j]=-2;break;}
                }
                ans[l++]=num+1;
            }
        }
        cout<<ans[0];
        for(int i=1;i<l;i++) cout<<" "<<ans[i];cout<<endl;
    }
    return 0;
}
/*
这样的S能用两种方法来表示（编码）：
1。用一个整数序列P＝p1p2...pn,其中pi是第i个右括号前的左括号个数
2。用一个整数序列W＝w1w2...wn，其中 wi 是从第 i 个右括号往左数直到遇到
和它相对应的左括号时经过的左括号个数（包括与第i个右括号相对应的左括号）。
比如：
S   ( ( ( () () () ) ) )
P: 4 5 6 6 6 6
W: 1 1 1 4 5 6
我们的任务就是将一个p序列转化为w序列
*/