Parencodings 1016

Parencodings

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Let S = s1 s2 ... s2n be a well-formed string of parentheses. S can be encoded in two different ways:

• 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).
• 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 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

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Source: Asia 2001, Tehran (Iran)

//思路是用0表示左括号，1表示右括号，再用几个for就出来了

#include <stdio.h>#include <string.h>int c[22];int n;void initP(){int x[n*3];int i,j;int k=0;int mi;int temp = c[0];while (temp--){x[k++] = 0;}x[k++] = 1;for (i=0; i<n; i++){if (c[i+1]!=-1){mi = c[i+1]-c[i];}for (j=0; j<mi; j++){x[k++] = 0;}x[k++] = 1;}int s[3*n];int e=0;int f[3*n];for (i=0; i<3*n; i++)s[i]=1;memset(f, 0, sizeof(f));int r;for (i=0; i<k-1; i++){if (x[i] == 1){for (j=i-1; j>=0; j--){if (x[j]==0 && f[j]!=1){for (r=j; r<i; r++)if (x[r]==1)(s[e])++;f[j]=1;break;}}e++;}}printf("%d", s[0]);for (i=1; i<e; i++){printf(" %d", s[i]);}printf("\n");}int main(){int N;scanf("%d", &N);int i;while (N--){memset(c, -1, sizeof(c));scanf("%d", &n);for (i=0; i<n; i++)scanf("%d", &c[i]);initP();}return 0;}