"""# ### 谜题描述 This is a harder version of the problem. In this version, n ≤ 300 000. Vasya is an experienced developer of programming competitions' problems. As all great minds at some time, Vasya faced a creative crisis. To improve the situation, Petya gifted him a string consisting of opening and closing brackets only. Petya believes, that the beauty of the bracket string is a number of its cyclical shifts, which form a correct bracket sequence. To digress from his problems, Vasya decided to select two positions of the string (not necessarily distinct) and swap characters located at this positions with each other. Vasya will apply this operation exactly once. He is curious what is the maximum possible beauty he can achieve this way. Please help him. We remind that bracket sequence s is called correct if: * s is empty; * s is equal to \"(t)\", where t is correct bracket sequence; * s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences. For example, \"(()())\", \"()\" are correct, while \")(\" and \"())\" are not. The cyclical shift of the string s of length n by k (0 ≤ k < n) is a string formed by a concatenation of the last k symbols of the string s with the first n - k symbols of string s. For example, the cyclical shift of string \"(())()\" by 2 equals \"()(())\". Cyclical shifts i and j are considered different, if i ≠ j. Input The first line contains an integer n (1 ≤ n ≤ 300 000), the length of the string. The second line contains a string, consisting of exactly n characters, where each of the characters is either \"(\" or \")\". Output The first line should contain a single integer — the largest beauty of the string, which can be achieved by swapping some two characters. The second line should contain integers l and r (1 ≤ l, r ≤ n) — the indices of two characters, which should be swapped in order to maximize the string's beauty. In case there are several possible swaps, print any of them. Examples Input 10 ()()())(() Output 5 8 7 Input 12 )(()(()())() Output 4 5 10 Input 6 )))(() Output 0 1 1 Note In the first example, we can swap 7-th and 8-th character, obtaining a string \"()()()()()\". The cyclical shifts by 0, 2, 4, 6, 8 of this string form a correct bracket sequence. In the second example, after swapping 5-th and 10-th character, we obtain a string \")(())()()(()\". The cyclical shifts by 11, 7, 5, 3 of this string form a correct bracket sequence. In the third example, swap of any two brackets results in 0 cyclical shifts being correct bracket sequences. Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution. ```python n=input() s=raw_input() a=[] maxx=0 for i in s: a.append(i) if i=='(': maxx+=1 else: maxx-=1 if maxx!=0: print 0 print 1,1 else: x=1 y=1 maxx=0 dp=[0]*n val=0 for i in range(n): if a[i]=='(': val+=1 else: val-=1 dp[i]=val minn=min(dp) for i in dp: if i==minn: maxx+=1 for i in range(n): for j in range(i,n): if(a[i]==a[j]): continue a[i],a[j]=a[j],a[i] dp=[0]*n val=0 for i1 in range(n): if a[i1]=='(': val+=1 else: val-=1 dp[i1]=val minn=min(dp) for i1 in dp: if i1==minn: val+=1 if val>maxx: maxx=val x=i+1 y=j+1 a[i],a[j]=a[j],a[i] print maxx print x,y ``` 请完成上述谜题的训练场环境类实现，包括所有必要的方法。 """ from bootcamp import Basebootcamp import random import re from bootcamp import Basebootcamp def compute_min_balance_and_count(s): balance = 0 min_balance = 0 count = 0 prefix = [] for c in s: balance += 1 if c == '(' else -1 prefix.append(balance) if balance < min_balance: min_balance = balance count = 1 elif balance == min_balance: count += 1 return min_balance, count, prefix def calculate_real_beauty(s): total = sum(1 if c == '(' else -1 for c in s) if total != 0: return 0 min_balance, count, prefix = compute_min_balance_and_count(s) overall_min = min(prefix) if overall_min < 0: return 0 return count def optimal_solution(n, s): max_beauty = 0 best_pair = (1, 1) original_beauty = calculate_real_beauty(s) max_beauty = original_beauty s_list = list(s) for i in range(n): for j in range(i, n): if s_list[i] == s_list[j]: continue # Perform swap s_list[i], s_list[j] = s_list[j], s_list[i] new_s = ''.join(s_list) current_beauty = calculate_real_beauty(new_s) if current_beauty > max_beauty: max_beauty = current_beauty best_pair = (i+1, j+1) # Revert swap s_list[i], s_list[j] = s_list[j], s_list[i] return (max_beauty, best_pair[0], best_pair[1]) class Btheworldisjustaprogrammingtaskhardversionbootcamp(Basebootcamp): def __init__(self, max_n=12): self.max_n = max_n # 控制案例规模保证验证效率 def case_generator(self): n = random.randint(1, self.max_n) # 生成有效测试案例（包含平衡和非平衡情况） if random.random() < 0.5 and n % 2 == 0: # 生成平衡括号字符串 s = ['(']*(n//2) + [')']*(n//2) random.shuffle(s) s = ''.join(s) else: # 随机生成可能不平衡的字符串 s = ''.join(random.choices(['(', ')'], k=n)) max_beauty, l, r = optimal_solution(n, s) return { 'n': n, 's': s, 'expected_max': max_beauty, 'swap_l': l, 'swap_r': r } @staticmethod def prompt_func(question_case): n = question_case['n'] s = question_case['s'] return f"""给定一个长度为{n}的括号字符串："{s}" 请通过交换两个字符（允许相同位置），最大化循环移位构成有效括号序列的数量。输出最大数量及交换位置（1-based）。 有效括号序列定义： 1. 空字符串 2. (A) 其中A是有效序列 3. AB 其中A和B都是有效序列 答案格式： [answer] {{最大数量}} {{位置1}} {{位置2}} [/answer]""" @staticmethod def extract_output(output): matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL) if not matches: return None content = matches[-1].strip() lines = [line.strip() for line in content.split('\n') if line.strip()] try: if len(lines) >= 2: max_val = int(lines[0]) l, r = map(int, lines[1].split()) return (max_val, l, r) except: pass return None @classmethod def _verify_correction(cls, solution, identity): if not solution or len(solution) != 3: return False max_val, l, r = solution # 验证基础参数 if max_val != identity['expected_max']: return False if not (1 <= l <= identity['n'] and 1 <= r <= identity['n']): return False # 执行交换操作 s_list = list(identity['s']) l_idx, r_idx = l-1, r-1 s_list[l_idx], s_list[r_idx] = s_list[r_idx], s_list[l_idx] new_s = ''.join(s_list) # 计算实际美丽值 actual_beauty = calculate_real_beauty(new_s) # 允许误差处理（应对计算误差） return actual_beauty == max_val