Parencodings

Problem 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

 

题意：一行括号数列，分别给出第i个右括号前有多少个左括号；要求求出与第i个右括号相对应的左括号所形成的括号对中包括多少左括号（包括本身与第i个右括号对应的左括号）

输入：

第一行：输入一个数t，表示有t组数列需要处理；接下来的2*t行，没两行是一组需要处理的数据，上一行第n表示有n个右括号；接下来的一行有n个整数；

AC代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>



using namespace std;


int main()
{
    freopen("1.txt","r",stdin);
    int dp[50]={0};//模拟数组，初始值都为0；
    int n,num,i;
    int s,j;
    int sum;
    cin>>num;//输入样例的个数；
    while(num)
    {

        cin>>n;//输入右括号的个数；
        for(i=0;i<n;i++){//输入值右括号前的左括号个数，并找打右括号的位置，将模拟数组的值付为1；
            cin>>j;
            dp[j+i]=1;
        }
        s=n;
        n<<=1;
        i=0;
        while(s){
            while(dp[i++]<=0);
            sum=0;
            for(j=i-2;j>=0;j--)
            {
                if(dp[j]<=0)sum++;//确定了一个左括号；
                if(dp[j]==0){dp[j]=-1;break;}//到了与相应右括号对应的左括号处，跳出循环。
            }
            cout<<sum<<" ";//输出左括号的个数。
            s--;
        }
        cout<<endl;
        memset(dp,0,sizeof(dp));//将dp数组制为0；
        num--;
    }
    return 0;
}