poj 1068--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

p-代表的是这个右括号前面有几个左括号
w-代表的是和这个右括号匹配的左括号在第几位
现在已知p矩阵，要求w的值。。
做完看到网上好多推出什么公式的。。我是直接用数组模拟的，先将所有的括号形式都存到数组中，然后在从后往前找与之匹配的，算出w的值，不算很好的方法，但是感觉应该好理解些。。

#include <stdio.h>
#include <string.h>
int main()
{
    int t,n;
    int i,j,k;
    int p[22],w[22],s[100];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0; i<n; i++)
            scanf("%d",&p[i]);
        for(i=0; i<p[0]; i++)
            s[i]=-1;
        s[i++]=1;
        for(j=1; j<n; j++)
        {
            for(k=0; k<p[j]-p[j-1]; k++)
                s[i++]=-1;
            s[i++]=1;
        }
        //for(j=0;j<i;j++)
        //printf("%d",s[j]);
        i=n-1;
        for(j=2*n-1; j>=0; j--)
        {
            if(s[j]==1)
            {
                int sum=0;
                for(k=j;; k--)
                {
                    sum+=s[k];
                    if(sum==0)
                        break;
                }
                w[i--]=(j-k+1)/2;
            }
        }
        printf("%d",w[0]);
        for(j=1; j<n; j++)
            printf(" %d",w[j]);
        printf("\n");
    }
    return 0;
}