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internbootcamp/bootcamp/ccircularrmq/ccircularrmq.py

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+"""# 
+
+### 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 <bits/stdc++.h>
+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']