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模拟括号

2024-07-05 15:11:40   226人阅读

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
#include<iostream>
#include<vector>
#include<cmath>
using namespace std;
int a[25],pp[1005];
int main()
{
    int t,n;
    cin>>t;
    vector<int> ve;
    vector<int>::iterator it;
    while(t--)
    {
        cin>>n;
        ve.clear();
        for(int i=0;i<n;i++)
            cin>>a[i];
        for(int i=0;i<a[0];i++)
        {
            ve.push_back(1);
        }
        ve.push_back(0);
        for(int i=1;i<n;i++)
        {
            for(int j=0;j<a[i]-a[i-1];j++)
            {
                ve.push_back(1);
            }
            ve.push_back(0);
        }
        int num=0;
        for(it=ve.begin();it!=ve.end();it++)
        {
             pp[num]=*it;
             num++;
        } 
        int sum=0;
        while(1)
        {
            int res=0;
            if(sum==num)
            {
                cout<<endl;
                break;
            }
            for(int i=0;i<num;i++)
            {
               for(int j=i+1;j<num;j++)
               {
                   if(pp[j]==-1)
                       continue;
                   if(pp[i]+pp[j]==1)
                   {
                       int temp=0;
                       for(int k=i;k<=j;k++)
                       {
                           temp+=pp[k];
                       } 
                       if(temp==1)
                           cout<<1<<" ";
                       else 
                           cout<<1+(abs(temp)+1)/2<<" "; 
                       pp[i]=-1;
                       pp[j]=-1;
                       sum+=2;
                       res=1;
                   }
                   else
                       break;
               } 
               if(res==1)
                   break;    
            }
        }
    }
    //system("pause");
    return 0;
}
View Code

 


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