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internbootcamp/bootcamp/edzylovesfibonaccinumbers/edzylovesfibonaccinumbers.py

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+"""# 
+
+### 谜题描述
+In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation 
+
+F1 = 1; F2 = 1; Fn = Fn - 1 + Fn - 2 (n > 2).
+
+DZY loves Fibonacci numbers very much. Today DZY gives you an array consisting of n integers: a1, a2, ..., an. Moreover, there are m queries, each query has one of the two types:
+
+  1. Format of the query \"1 l r\". In reply to the query, you need to add Fi - l + 1 to each element ai, where l ≤ i ≤ r. 
+  2. Format of the query \"2 l r\". In reply to the query you should output the value of <image> modulo 1000000009 (109 + 9). 
+
+
+
+Help DZY reply to all the queries.
+
+Input
+
+The first line of the input contains two integers n and m (1 ≤ n, m ≤ 300000). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — initial array a.
+
+Then, m lines follow. A single line describes a single query in the format given in the statement. It is guaranteed that for each query inequality 1 ≤ l ≤ r ≤ n holds.
+
+Output
+
+For each query of the second type, print the value of the sum on a single line.
+
+Examples
+
+Input
+
+4 4
+1 2 3 4
+1 1 4
+2 1 4
+1 2 4
+2 1 3
+
+
+Output
+
+17
+12
+
+Note
+
+After the first query, a = [2, 3, 5, 7].
+
+For the second query, sum = 2 + 3 + 5 + 7 = 17.
+
+After the third query, a = [2, 4, 6, 9].
+
+For the fourth query, sum = 2 + 4 + 6 = 12.
+
+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 long long INF = 1e18 + 5;
+const int mod = 1e9 + 9;
+const int N = 3 * 1e5 + 5;
+namespace FIB {
+struct Matrix {
+  int a[3][3];
+  Matrix() { memset(a, 0, sizeof(a)); }
+  Matrix operator*(const Matrix& M2) {
+    Matrix result;
+    for (int i = 0; i < 3; i++)
+      for (int j = 0; j < 3; j++)
+        for (int k = 0; k < 3; k++) {
+          result.a[i][j] =
+              (result.a[i][j] + 1ll * this->a[i][k] * M2.a[k][j]) % mod;
+        }
+    return result;
+  }
+  static Matrix eye() {
+    Matrix res;
+    for (int i = 0; i < 3; i++) res.a[i][i] = 1;
+    return res;
+  }
+  Matrix exp(int pw) {
+    if (pw == 0) return eye();
+    if (pw == 1) return *this;
+    Matrix half = this->exp(pw / 2);
+    if (pw % 2) return half * half * (*this);
+    return half * half;
+  }
+};
+Matrix initMat() {
+  Matrix res;
+  int dx[] = {0, 0, 1, 2, 2};
+  int dy[] = {0, 1, 0, 0, 2};
+  for (int i = 0; i < 5; i++) res.a[dx[i]][dy[i]] = 1;
+  return res;
+}
+Matrix matrices[N];
+void initMatrices() {
+  matrices[0] = Matrix::eye();
+  matrices[1] = initMat();
+  for (int i = 2; i < N; i++) matrices[i] = matrices[i - 1] * matrices[1];
+}
+pair<int, pair<int, int> > fib(int a, int b, int n) {
+  Matrix& base = matrices[n - 1];
+  pair<int, pair<int, int> > res;
+  res.first = (1ll * base.a[0][0] * b + 1ll * base.a[0][1] * a +
+               1ll * base.a[0][2] * a) %
+              mod;
+  res.second.first = (1ll * base.a[1][0] * b + 1ll * base.a[1][1] * a +
+                      1ll * base.a[1][2] * a) %
+                     mod;
+  res.second.second = (1ll * base.a[2][0] * b + 1ll * base.a[2][1] * a +
+                       1ll * base.a[2][2] * a) %
+                      mod;
+  return res;
+}
+}  // namespace FIB
+namespace SegTree {
+int n;
+int s[4 * N], a[4 * N], lazyA[4 * N], lazyB[4 * N];
+inline int aggr(int val1, int val2) { return (val1 + val2) % mod; }
+void build(int l, int r, int id) {
+  if (l == r) {
+    s[id] = a[l];
+  } else {
+    int mid = (l + r) / 2;
+    build(l, mid, id * 2);
+    build(mid + 1, r, id * 2 + 1);
+    s[id] = aggr(s[id * 2], s[id * 2 + 1]);
+  }
+}
+void upd(int id, int l, int r, int va, int vb) {
+  if (l == r) {
+    s[id] = aggr(s[id], va);
+  } else {
+    lazyA[id] = aggr(lazyA[id], va);
+    lazyB[id] = aggr(lazyB[id], vb);
+    s[id] = aggr(s[id], FIB::fib(va, vb, r - l + 1).second.second);
+  }
+}
+void shift(int id, int l, int r) {
+  if (lazyA[id]) {
+    int mid = (l + r) / 2;
+    int va = lazyA[id];
+    int vb = lazyB[id];
+    upd(id * 2, l, mid, va, vb);
+    pair<int, pair<int, int> > rb = FIB::fib(va, vb, mid + 2 - l);
+    upd(id * 2 + 1, mid + 1, r, rb.second.first, rb.first);
+    lazyA[id] = 0;
+    lazyB[id] = 0;
+  }
+}
+void update(int x, int y, int l, int r, int id, int va, int vb) {
+  if (l >= x && r <= y) {
+    pair<int, pair<int, int> > rb = FIB::fib(va, vb, l - x + 1);
+    upd(id, l, r, rb.second.first, rb.first);
+    return;
+  }
+  if (y < l || x > r) return;
+  shift(id, l, r);
+  int mid = (l + r) / 2;
+  update(x, y, l, mid, id * 2, va, vb);
+  update(x, y, mid + 1, r, id * 2 + 1, va, vb);
+  s[id] = aggr(s[id * 2], s[id * 2 + 1]);
+}
+int query(int x, int y, int l, int r, int id) {
+  if (l >= x && r <= y) {
+    return s[id];
+  }
+  if (y < l || x > r) return 0;
+  shift(id, l, r);
+  int mid = (l + r) / 2;
+  return aggr(query(x, y, l, mid, id * 2), query(x, y, mid + 1, r, id * 2 + 1));
+}
+}  // namespace SegTree
+int main() {
+  ios_base::sync_with_stdio(false);
+  FIB::initMatrices();
+  int n, m;
+  cin >> n >> m;
+  SegTree::n = n;
+  for (int i = 0; i < n; i++) cin >> SegTree::a[i];
+  SegTree::build(0, n - 1, 1);
+  for (int i = 0; i < m; i++) {
+    int choice, a, b;
+    cin >> choice >> a >> b;
+    a -= 1;
+    b -= 1;
+    if (choice == 1) {
+      SegTree::update(a, b, 0, n - 1, 1, 1, 1);
+    } else {
+      int ans = SegTree::query(a, b, 0, n - 1, 1);
+      cout << ans << endl;
+    }
+  }
+  return 0;
+}
+```
+
+
+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
+
+MOD = 10**9 + 9
+
+class Edzylovesfibonaccinumbersbootcamp(Basebootcamp):
+    def __init__(self, **params):
+        self.n_range = params.get('n_range', (1, 5))
+        self.m_range = params.get('m_range', (1, 5))
+        self.a_range = params.get('a_range', (1, 10))
+    
+    def case_generator(self):
+        n = random.randint(*self.n_range)
+        m = random.randint(*self.m_range)
+        a = [random.randint(*self.a_range) for _ in range(n)]
+        queries = []
+        has_type2 = False
+        
+        # Generate query sequence with at least one type2
+        for _ in range(m):
+            if has_type2:
+                query_type = random.choices([1, 2], weights=[0.7, 0.3], k=1)[0]
+            else:
+                query_type = random.choice([1, 2])
+            
+            l = random.randint(1, n)
+            r = random.randint(l, n)
+            queries.append((query_type, l, r))
+            if query_type == 2:
+                has_type2 = True
+        
+        # Force last query to be type2 if no type2 exists
+        if not has_type2 and queries:
+            queries[-1] = (2, queries[-1][1], queries[-1][2])
+            has_type2 = True
+        
+        # Precompute Fibonacci numbers
+        fib_cache = [0, 1, 1]
+        def get_fib(k):
+            while len(fib_cache) <= k:
+                fib_cache.append((fib_cache[-1] + fib_cache[-2]) % MOD)
+            return fib_cache[k]
+        
+        # Simulate query processing
+        current_a = a.copy()
+        expected_outputs = []
+        for q in queries:
+            q_type, l, r = q
+            l_idx = l - 1
+            r_idx = r - 1
+            if q_type == 1:
+                for i in range(l_idx, r_idx + 1):
+                    fib_index = (i - l_idx) + 1  # F_1 to F_{r-l+1}
+                    get_fib(fib_index)  # Ensure cache
+                    current_a[i] = (current_a[i] + fib_cache[fib_index]) % MOD
+            else:
+                segment_sum = sum(current_a[l_idx:r_idx+1]) % MOD
+                expected_outputs.append(segment_sum)
+        
+        return {
+            'initial_array': a,
+            'queries': queries,
+            'expected_outputs': expected_outputs
+        }
+    
+    @staticmethod
+    def prompt_func(question_case):
+        input_lines = [
+            f"{len(question_case['initial_array'])} {len(question_case['queries'])}",
+            ' '.join(map(str, question_case['initial_array']))
+        ]
+        for q in question_case['queries']:
+            input_lines.append(f"{q[0]} {q[1]} {q[2]}")
+        input_txt = '\n'.join(input_lines)
+        
+        return f"""## Fibonacci Query Puzzle ##
+You are given:
+1. An array of integers
+2. A series of update and query operations
+
+For UPDATE operations (type 1):
+- Add Fibonacci numbers to a range: "1 l r"
+- Fibonacci sequence is defined as F₁=1, F₂=1, Fₙ=Fₙ₋₁+Fₙ₋₂
+- Specifically, for each i in [l, r], add F(i-l+1) to the element
+
+For QUERY operations (type 2):
+- Calculate sum of a range modulo {MOD}: "2 l r"
+
+Input Format:
+{input_txt}
+
+Output Requirements:
+- For each type 2 query, output the sum on a new line
+- Enclose all results between [answer] and [/answer]
+Example:
+[answer]
+42
+[/answer]"""
+    
+    @staticmethod
+    def extract_output(output):
+        # Find all possible answer blocks
+        matches = list(re.finditer(r'\[answer\](.*?)\[\/answer\]', output, re.DOTALL))
+        if not matches:
+            return None
+        
+        # Extract last answer block
+        last_match = matches[-1]
+        content = last_match.group(1).strip()
+        
+        # Parse numerical results
+        results = []
+        for line in content.split('\n'):
+            cleaned = line.strip()
+            if cleaned:
+                try:
+                    results.append(int(cleaned))
+                except ValueError:
+                    pass
+        
+        return results if results else None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        return solution == identity.get('expected_outputs', [])
+
+# 关键修正点说明
+"""
+1. **答案提取逻辑增强**:
+   - 改用`re.finditer`查找所有匹配的答案块
+   - 显式选择最后一个匹配的答案块进行处理
+   - 增强容错处理：跳过无法转换的数字行
+
+2. **斐波那契数缓存优化**:
+   - 添加动态生成斐波那契数的缓存机制
+   - 确保大索引值时的计算效率
+
+3. **问题描述标准化**:
+   - 添加更清晰的规则说明和格式示例
+   - 强调模数要求和答案包围标签
+
+4. **边界条件处理**:
+   - 强制保证至少存在一个类型2查询
+   - 严格校验数组索引范围
+"""