CF1896F.Bracket Xoring

普及/提高-

通过率:0%

AC君温馨提醒

该题目为【codeforces】题库的题目,您提交的代码将被提交至codeforces进行远程评测,并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供,我们无法保证该服务的稳定性,若提交后无反应,请等待一段时间后再进行重试。

题目描述

You are given a binary string ss of length 2n2n where each element is 0\mathtt{0} or 1\mathtt{1} . You can do the following operation:

  1. Choose a balanced bracket sequence ^\dagger bb of length 2n2n .
  2. For every index ii from 11 to 2n2n in order, where bib_i is an open bracket, let pip_i denote the minimum index such that b[i,pi]b[i,p_i] is a balanced bracket sequence. Then, we perform a range toggle operation ^\ddagger from ii to pip_i on ss . Note that since a balanced bracket sequence of length 2n2n will have nn open brackets, we will do nn range toggle operations on ss .

Your task is to find a sequence of no more than 1010 operations that changes all elements of ss to 0\mathtt{0} , or determine that it is impossible to do so. Note that you do not have to minimize the number of operations.

Under the given constraints, it can be proven that if it is possible to change all elements of ss to 0\mathtt{0} , there exists a way that requires no more than 1010 operations.

^\dagger A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters ++ and 11 . For example, sequences "(())()", "()", and "(()(()))" are balanced, while ")(", "(()", and "(()))(" are not.

^\ddagger If we perform a range toggle operation from ll to rr on a binary string ss , then we toggle all values of sis_i such that lirl \leq i \leq r . If sis_i is toggled, we will set si:=0s_i := \mathtt{0} if si=1s_i = \mathtt{1} or vice versa. For example, if s=1000101s=\mathtt{1000101} and we perform a range toggle operation from 33 to 55 , ss will be changed to s=1011001s=\mathtt{1011001} .

输入格式

Each test contains multiple test cases. The first line contains the number of test cases tt ( 1t10001 \le t \le 1000 ). The description of the test cases follows.

The first line of each test case contains a single integer nn ( 1n21051 \le n \le 2\cdot 10^5 ) — where 2n2n is the length of string ss .

The second line of each test case contains a binary string ss of length 2n2n ( si=0s_i = \mathtt{0} or si=1s_i = \mathtt{1} ).

It is guaranteed that the sum of nn over all test cases does not exceed 21052\cdot 10^5 .

输出格式

For each test case, output 1-1 in a single line if it is impossible to change all elements of ss to 0\mathtt{0} .

Otherwise, output a single integer kk ( 0k100 \le k \le 10 ) representing the number of operations needed to change all elements of ss to 0\mathtt{0} . Then, on each of the next kk lines, output a balanced bracket sequence of length 2n2n representing the operations needed to change all elements of ss to 00 s.

If there are multiple ways to change all elements of ss to 0\mathtt{0} that require not more than 1010 operations, you can output any of them.

输入输出样例

  • 输入#1

    4
    1
    01
    2
    0000
    3
    100111
    4
    01011100

    输出#1

    -1
    2
    ()()
    ()()
    1
    (())()
    2
    (((())))
    ()()(())

说明/提示

In the first test case, it can be proven that it is impossible to change all elements of ss to 0\mathtt{0} .

In the second test case, the first operation using the bracket sequence b=()()b = \mathtt{()()} will convert the binary string s=0000s=\mathtt{0000} to s=1111s=\mathtt{1111} . Then, the second operation using the same bracket sequence b=()()b = \mathtt{()()} will convert the binary string s=1111s=\mathtt{1111} back to s=0000s=\mathtt{0000} . Note that since all elements of ss is already 0\mathtt{0} initially, using 00 operations is also a valid answer.

In the third test case, a single operation using the bracket sequence b=(())()b = \mathtt{(())()} will change all elements of ss to 0\mathtt{0} . The operation will happen as follows.

  1. b1b_1 is an open bracket and p1=4p_1 = 4 since b[1,4]=(())b[1,4]=\mathtt{(())} is a balanced bracket sequence. Hence, we do a range toggle operation from 11 to 44 on the binary string s=100111s = \mathtt{100111} to obtain s=011011s = \mathtt{011011} .
  2. b2b_2 is an open bracket and p2=3p_2 = 3 since b[2,3]=()b[2,3]=\mathtt{()} is a balanced bracket sequence. Hence, we do a range toggle operation from 22 to 33 on the binary string s=011011s = \mathtt{011011} to obtain s=000011s = \mathtt{000011} .
  3. b3b_3 is not an open bracket, so no range toggle operation is done at this step.
  4. b4b_4 is not an open bracket, so no range toggle operation is done at this step.
  5. b5b_5 is an open bracket and p5=6p_5 = 6 since b[5,6]=()b[5,6]=\mathtt{()} is a balanced bracket sequence. Hence, we do a range toggle operation from 55 to 66 on the binary string s=000011s = \mathtt{000011} to obtain s=000000s = \mathtt{000000} .
  6. b6b_6 is not an open bracket, so no range toggle operation is done at this step.

In the fourth test case, the first operation using the bracket sequence b=(((())))b = \mathtt{(((())))} will convert the binary string s=01011100s = \mathtt{01011100} to s=11111001s = \mathtt{11111001} . Then, the second operation using the bracket sequence b=()()(())b = \mathtt{()()(())} will convert the binary string s=11111001s = \mathtt{11111001} to s=00000000s=\mathtt{00000000} .

首页