A. Regular Bracket Sequences

题目如下：

A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not.

You are given an integer n

. Your goal is to construct and print exactly n different regular bracket sequences of length 2n

Input

The first line contains one integer t

(1≤t≤50) — the number of test cases.

Each test case consists of one line containing one integer n

(1≤n≤50).

Output

For each test case, print n

lines, each containing a regular bracket sequence of length exactly 2n. All bracket sequences you output for a testcase should be different (though they may repeat in different test cases). If there are multiple answers, print any of them. It can be shown that it's always possible.

AC代码如下：

#include<stdio.h>
int main()
{int t;int n;scanf("%d", &t);while (t--){scanf("%d", &n);int tmp = n;int i;while (tmp--){for (i = 1; i <= tmp; i++){printf("(");}for (i = 1; i <= n - tmp; i++){printf("()");}for (i = 1; i <= tmp; i++){printf(")");}printf("\n");}}
}

代码解释：

核心————》找规律

：**‘*’**————》（）单独排列

（）里面加 *

（（））里面＋*

。。。。。。。。。。。。

按照这个规律我们就可以写代码了，如果仍不明白，请结合代码与文字再理解一下。

开学的第三天，努力努力——》我想打ACM！