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：

s是一组括弧。有两种编码方法。
方法一，可获得数列P。其中每个Pi是指在第i个右括弧左边的左括弧数目。(((()比如以上在第一个右括弧左边有4个左括弧。所以P1是4，p2是5...

方法二，可获得数列W。把这堆括弧左右一一对应起来。从第一个右括号数起，每个wi是指第i个右括号与对应的左括号之间的完整括号数，包括他本身

方法一：

#include<stdio.h>
#include<string.h>
int main()
{
	int i,l,j,a[1000],b[1000],q=0;
	char c[1000];
	memset(a,0,sizeof(a));
	memset(b,0,sizeof(b));
	memset(c,0,sizeof(c));//初始化
	int m,n,t;
	scanf("%d",&m);
	while(m--)
	{
		q=0;
		scanf("%d",&n);
		for(i=0;i<n;i++)
		    scanf("%d",&a[i]);
		//将p数列转化为括号字符串，存在字符数组C中
        for(i=0;i<a[0];i++)
           c[i]='(';//c[a[0]]左边的都是“(”
        c[a[0]]=')';
        t=a[0]+1;//继续下去转化
        for(i=1;i<n;i++)
        {
        	int s=a[i]-a[i-1];
        	for(j=t;j<t+s;j++)
        	    c[j]='(';
    	    t=t+s;
    	    c[t]=')';
    	    t=t+1;
		}
		j=0;
		int k=0,q;
		for(i=0;i<t;i++)
		{
			if(c[i]==')')
			{
				k=0;
				if(c[i-1]=='(')
				{
					c[i]='}';
					c[i-1]='{';
					k=1;
				}//当左右括号相邻时，所包含的括号数是1，并将找过的括号用“{ ”，“}”标记
				else
				{
					c[i]='}';
					for(q=i-1;;q--)
					{
						if(c[q]=='{')
							k=k+1;
						else if(c[q]=='(')
						{
							k=k+1;
							c[q]='{';
							break;
						}
						else
							continue;
					}//从遇到')'的位置往后找，遇到标记过的，括号数加1，直到遇到')'为止
				}
				b[j++]=k;//将得到的括号数存入数组
			}
		}
		for(i=0;i<j-1;i++)
			printf("%d ",b[i]);
		printf("%d\n",b[j-1]);
	} 
	return 0;
}

方法二：

#include<stdio.h>
int main()
{
    int count;
    int n,j,i,a[1000],w[1000],p[1000];
    scanf("%d",&count);
    while(count--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++) scanf("%d",&p[i]);
        a[0]=p[1];
        for(i=1;i<n;i++) a[i]=p[i+1]-p[i];
        for(i=1;i<=n;i++)
        {
            for(j=i-1;j>=0;j--)
            {
                if(a[j]>0)
				{
					a[j]--;
					break;
				}
            }
            w[i]=i-j;
        }
        for(int i=1;i<n;i++)printf("%d ",w[i]);
        printf("%d\n",w[n]);
    }
    return 0;
}