Parencodings

I - Parencodings

Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:10000KB    64bit IO Format:%I64d & %I64u

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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 6666W-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

264 5 6 6 6 69 4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 61 1 2 4 5 1 1 3 9

先通过输入数据把括号字符数组模拟出来，再从第一个右括号开始向前找对应的左括号，找到后做上标记以防重复配对。

#include<stdio.h>int n[25]={0},w[25]={0},r[25];char s[25];void main(){int m,i,j,z,t,d,p;scanf("%d",&z);while(z--){scanf("%d",&m);for(i=1;i<=m;i++)scanf("%d",&n[i]);for(j=1;j<=n[1];j++)s[j]='(';for(i=2,s[j++]=')';i<m;i++)if(n[i]>n[i-1]){t=n[i]-n[i-1];while(t--)s[j++]='(';s[j++]=')';}else if(n[i]==n[i-1])s[j++]=')';for(i=1;i<=m;i++)            w[i]=1;for(i=1,d=0;i<=m;i++)            for(j=n[i];j>=0;j--)            if(w[j])            {                r[d++]=n[i]-j+1;                w[j]=0;break;            }        for(i=0;i<d;i++)if(i!=d-1)printf("%d ",r[i]);else printf("%d\n",r[i]);}}