"""# ### 谜题描述 You are given circular array a0, a1, ..., an - 1. There are two types of operations with it: * inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v; * rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively). Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1. Write program to process given sequence of operations. Input The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 106 ≤ v ≤ 106) — inc operation. Output For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d). Examples Input 4 1 2 3 4 4 3 0 3 0 -1 0 1 2 1 Output 1 0 0 Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution. ```python #include using namespace std; const int N = 2e5 + 5; struct segNode { long long val; long long lazy; int l, r; segNode() { val = lazy = 0; } segNode(int a, int b, int c) { val = a; l = b; r = c; lazy = 0; } }; int arr[N]; segNode segTree[4 * N]; void shift(int num, int left, int right) { int upd = segTree[num].lazy; segTree[num].val += upd; if (segTree[num].l != segTree[num].r) { segTree[left].lazy += upd; segTree[right].lazy += upd; } segTree[num].lazy = 0; } void build(int l, int r, int num) { if (l > r) return; if (l == r) { segTree[num] = segNode(arr[l], l, r); return; } int mid = (l + r) >> 1, left = num << 1, right = left | 1; build(l, mid, left); build(mid + 1, r, right); segTree[num] = segNode(min(segTree[left].val, segTree[right].val), l, r); } void update(int l, int r, int num, int lu, int ru, int inc) { if (l > r) return; int mid = (l + r) >> 1, left = num << 1, right = left | 1; if (segTree[num].lazy != 0) shift(num, left, right); if (l > ru || r < lu) return; if (l >= lu && r <= ru) { segTree[num].lazy += inc; shift(num, left, right); return; } update(l, mid, left, lu, ru, inc); update(mid + 1, r, right, lu, ru, inc); segTree[num].val = min(segTree[right].val, segTree[left].val); } long long query(int l, int r, int num, int lq, int rq) { if (l > r) return (long long)1e17; int mid = (l + r) >> 1, left = num << 1, right = left | 1; if (segTree[num].lazy != 0) shift(num, left, right); if (l > rq || r < lq) return (long long)1e17; if (l >= lq && r <= rq) { return segTree[num].val; } return min(query(l, mid, left, lq, rq), query(mid + 1, r, right, lq, rq)); } int main() { int n; scanf(\"%d\", &n); for (int i = 0; i < n; ++i) { scanf(\"%d\", arr + i); } build(0, n - 1, 1); int q; scanf(\"%d\", &q); while (q--) { int x, y; char ch; scanf(\"%d%d%c\", &x, &y, &ch); if (ch == ' ') { int v; scanf(\"%d\", &v); if (x <= y) update(0, n - 1, 1, x, y, v); else update(0, n - 1, 1, x, n - 1, v), update(0, n - 1, 1, 0, y, v); } else { long long ans; if (x <= y) ans = query(0, n - 1, 1, x, y); else ans = min(query(0, n - 1, 1, x, n - 1), query(0, n - 1, 1, 0, y)); printf(\"%lld\n\", ans); } } } ``` 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 class Ccircularrmqbootcamp(Basebootcamp): def __init__(self, n_range=(4, 10), m_range=(3, 8), **params): self.n_range = n_range self.m_range = m_range super().__init__(**params) def case_generator(self): n = random.randint(*self.n_range) array = [random.randint(-10, 10) for _ in range(n)] m = random.randint(*self.m_range) operations = [] expected_outputs = [] current_array = array.copy() for _ in range(m): if random.random() < 0.5: # Generate inc operation lf = random.randint(0, n-1) rg = random.randint(0, n-1) v = random.randint(-5, 5) operations.append({'type': 'inc', 'lf': lf, 'rg': rg, 'v': v}) # Apply inc to current_array if lf <= rg: for i in range(lf, rg+1): current_array[i] += v else: for i in range(lf, n): current_array[i] += v for i in range(0, rg+1): current_array[i] += v else: # Generate rmq operation lf = random.randint(0, n-1) rg = random.randint(0, n-1) operations.append({'type': 'rmq', 'lf': lf, 'rg': rg}) # Calculate min value min_val = float('inf') if lf <= rg: for i in range(lf, rg+1): if current_array[i] < min_val: min_val = current_array[i] else: for i in range(lf, n): if current_array[i] < min_val: min_val = current_array[i] for i in range(0, rg+1): if current_array[i] < min_val: min_val = current_array[i] expected_outputs.append(min_val) return { 'n': n, 'array': array, 'operations': operations, 'expected_outputs': expected_outputs } @staticmethod def prompt_func(question_case) -> str: n = question_case['n'] array_str = ' '.join(map(str, question_case['array'])) m = len(question_case['operations']) input_lines = [str(n), array_str, str(m)] for op in question_case['operations']: if op['type'] == 'inc': line = f"{op['lf']} {op['rg']} {op['v']}" else: line = f"{op['lf']} {op['rg']}" input_lines.append(line) input_data = '\n'.join(input_lines) prompt = f"""你是一个编程竞赛的参赛者，需要解决以下问题： 给定一个长度为{n}的循环数组，处理一系列操作并输出结果。数组初始值为：{array_str}。 操作类型： 1. inc lf rg v：将循环区间[lf, rg]内的每个元素增加v。当lf <= rg时，区间是连续的；否则包含数组末尾到开头。 2. rmq lf rg：查询循环区间[lf, rg]内的最小值。 输入格式： {input_data} 请处理所有操作，并按顺序输出每个rmq操作的结果。将结果按顺序放在[answer]和[/answer]标签之间，每个结果占一行。 例如，若结果应为1、0、0，则正确格式为： [answer] 1 0 0 [/answer]""" return prompt @staticmethod def extract_output(output): pattern = re.compile(r'\[answer\](.*?)\[\/answer\]', re.DOTALL) matches = pattern.findall(output) if not matches: return None content = matches[-1].strip() solutions = [] for line in content.split('\n'): line = line.strip() if line: try: solutions.append(int(line)) except: return None return solutions @classmethod def _verify_correction(cls, solution, identity): return solution == identity['expected_outputs']