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internbootcamp/bootcamp/eonchangingtree/eonchangingtree.py

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+"""# 
+
+### 谜题描述
+You are given a rooted tree consisting of n vertices numbered from 1 to n. The root of the tree is a vertex number 1.
+
+Initially all vertices contain number 0. Then come q queries, each query has one of the two types:
+
+  * The format of the query: 1 v x k. In response to the query, you need to add to the number at vertex v number x; to the numbers at the descendants of vertex v at distance 1, add x - k; and so on, to the numbers written in the descendants of vertex v at distance i, you need to add x - (i·k). The distance between two vertices is the number of edges in the shortest path between these vertices. 
+  * The format of the query: 2 v. In reply to the query you should print the number written in vertex v modulo 1000000007 (109 + 7). 
+
+
+
+Process the queries given in the input.
+
+Input
+
+The first line contains integer n (1 ≤ n ≤ 3·105) — the number of vertices in the tree. The second line contains n - 1 integers p2, p3, ... pn (1 ≤ pi < i), where pi is the number of the vertex that is the parent of vertex i in the tree.
+
+The third line contains integer q (1 ≤ q ≤ 3·105) — the number of queries. Next q lines contain the queries, one per line. The first number in the line is type. It represents the type of the query. If type = 1, then next follow space-separated integers v, x, k (1 ≤ v ≤ n; 0 ≤ x < 109 + 7; 0 ≤ k < 109 + 7). If type = 2, then next follows integer v (1 ≤ v ≤ n) — the vertex where you need to find the value of the number.
+
+Output
+
+For each query of the second type print on a single line the number written in the vertex from the query. Print the number modulo 1000000007 (109 + 7).
+
+Examples
+
+Input
+
+3
+1 1
+3
+1 1 2 1
+2 1
+2 2
+
+
+Output
+
+2
+1
+
+Note
+
+You can read about a rooted tree here: http://en.wikipedia.org/wiki/Tree_(graph_theory).
+
+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>
+const int N = 300200;
+using namespace std;
+int n;
+int tim;
+int tin[N];
+int tout[N];
+int dep[N];
+int mod = 1e9 + 7;
+pair<int, int> t[4 * N];
+vector<int> v[N];
+void dfs(int x, int g) {
+  tin[x] = tout[x] = ++tim;
+  dep[tin[x]] = g;
+  for (auto y : v[x]) {
+    dfs(y, g - 1);
+    tout[x] = tout[y];
+  }
+}
+void push(int x) {
+  t[x * 2].first += t[x].first;
+  t[x * 2].second += t[x].second;
+  t[x * 2 + 1].first += t[x].first;
+  t[x * 2 + 1].second += t[x].second;
+  t[x * 2].first %= mod;
+  t[x * 2].second %= mod;
+  t[x * 2 + 1].first %= mod;
+  t[x * 2 + 1].second %= mod;
+  t[x].first = t[x].second = 0;
+}
+void upd(int x, int l, int r, int tl, int tr, int f1, int f2) {
+  if (tl > tr) return;
+  if (l == tl && r == tr) {
+    t[x].first = (t[x].first + f1) % mod;
+    t[x].second = (t[x].second + f2) % mod;
+    return;
+  }
+  push(x);
+  int m = (l + r) / 2;
+  upd(x * 2, l, m, tl, min(m, tr), f1, f2);
+  upd(x * 2 + 1, m + 1, r, max(m + 1, tl), tr, f1, f2);
+}
+int get(int x, int l, int r, int g) {
+  if (l == r) {
+    int ans = (1ll * t[x].second * dep[l]) % mod;
+    ans = (t[x].first + ans) % mod;
+    return ans;
+  }
+  push(x);
+  int m = (l + r) / 2;
+  if (g <= m)
+    return get(x * 2, l, m, g);
+  else
+    return get(x * 2 + 1, m + 1, r, g);
+}
+int main() {
+  ios_base::sync_with_stdio(0);
+  scanf(\"%d\", &n);
+  for (int i = 2; i <= n; i++) {
+    int x;
+    scanf(\"%d\", &x);
+    v[x].push_back(i);
+  }
+  dfs(1, n);
+  int q;
+  scanf(\"%d\", &q);
+  for (int i = 1; i <= q; i++) {
+    int t, v, x, k;
+    scanf(\"%d\", &t);
+    if (t == 1) {
+      scanf(\"%d%d%d\", &v, &x, &k);
+      long long f = 1ll * x - 1ll * dep[tin[v]] * k;
+      f = f % mod + mod;
+      upd(1, 1, n, tin[v], tout[v], f % mod, k);
+    } else {
+      scanf(\"%d\", &v);
+      printf(\"%d\n\", get(1, 1, n, tin[v]));
+    }
+  }
+}
+```
+
+
+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 + 7
+
+def build_adj(parents, n):
+    adj = {i: [] for i in range(1, n+1)}
+    for i in range(2, n+1):
+        parent = parents[i-2]
+        adj[parent].append(i)
+    return adj
+
+def dfs(x, parent_adj, tin, tout, dep, current_time, current_g):
+    current_time[0] += 1
+    tin[x] = current_time[0]
+    dep[tin[x]] = current_g
+    for child in parent_adj.get(x, []):
+        dfs(child, parent_adj, tin, tout, dep, current_time, current_g-1)
+    tout[x] = current_time[0]
+
+def perform_dfs(n, parent_adj):
+    tin = [0] * (n + 1)
+    tout = [0] * (n + 1)
+    dep = [0] * (n + 2)  # tin values are 1-based
+    current_time = [0]
+    dfs(1, parent_adj, tin, tout, dep, current_time, n)
+    return tin, tout, dep
+
+def process_queries_for_identity(queries, n, tin_dict, tout_dict, dep_list):
+    a = [0] * (n + 2)
+    b = [0] * (n + 2)
+    expected_outputs = []
+    for query in queries:
+        if query['type'] == 1:
+            v = query['v']
+            x = query['x']
+            k = query['k']
+            tin_v = tin_dict[v]
+            tout_v = tout_dict[v]
+            f1 = (x - dep_list[tin_v] * k) % mod
+            f2 = k % mod
+            for u in range(1, n+1):
+                u_tin = tin_dict[u]
+                if tin_v <= u_tin <= tout_v:
+                    a[u_tin] = (a[u_tin] + f1) % mod
+                    b[u_tin] = (b[u_tin] + f2) % mod
+        else:
+            v = query['v']
+            u_tin = tin_dict[v]
+            res = (a[u_tin] + b[u_tin] * dep_list[u_tin]) % mod
+            expected_outputs.append(res)
+    return expected_outputs
+
+class Eonchangingtreebootcamp(Basebootcamp):
+    def __init__(self, max_n=5, max_q=5):
+        self.max_n = max_n
+        self.max_q = max_q
+    
+    def case_generator(self):
+        n = random.randint(1, self.max_n)
+        parents = []
+        if n > 1:
+            parents = [random.randint(1, i-1) for i in range(2, n+1)]
+        adj = build_adj(parents, n)
+        tin, tout, dep = perform_dfs(n, adj)
+        tin_dict = {x: tin[x] for x in range(1, n+1)}
+        tout_dict = {x: tout[x] for x in range(1, n+1)}
+        q = random.randint(1, self.max_q)
+        queries = []
+        for _ in range(q):
+            if random.random() < 0.3 or not any(q.get('type') == 2 for q in queries):
+                v = random.randint(1, n)
+                queries.append({'type': 2, 'v': v})
+            else:
+                v = random.randint(1, n)
+                x = random.randint(0, mod-1)
+                k = random.randint(0, mod-1)
+                queries.append({'type': 1, 'v': v, 'x': x, 'k': k})
+        expected_outputs = process_queries_for_identity(queries, n, tin_dict, tout_dict, dep)
+        new_queries = []
+        output_idx = 0
+        for q in queries:
+            if q['type'] == 2:
+                new_q = q.copy()
+                new_q['expected'] = expected_outputs[output_idx]
+                new_queries.append(new_q)
+                output_idx += 1
+            else:
+                new_queries.append(q)
+        return {
+            'n': n,
+            'parents': parents,
+            'queries': new_queries,
+            'tin_dict': tin_dict,
+            'tout_dict': tout_dict,
+            'dep_list': dep
+        }
+    
+    @staticmethod
+    def prompt_func(question_case):
+        input_lines = [
+            str(question_case['n']),
+            ' '.join(map(str, question_case['parents'])) if question_case['n'] > 1 else '',
+            str(len(question_case['queries']))
+        ]
+        for q in question_case['queries']:
+            if q['type'] == 1:
+                input_lines.append(f"1 {q['v']} {q['x']} {q['k']}")
+            else:
+                input_lines.append(f"2 {q['v']}")
+        input_str = '\n'.join(input_lines)
+        return f"""你正在解决一个树处理问题。处理所有查询并将每个类型2的答案放在[answer]和[/answer]之间。输入数据如下：
+
+{input_str}
+
+规则：
+1. 类型1查询为节点v及其后代按距离加值。
+2. 类型2查询输出节点值模10^9+7的结果。
+
+答案格式示例：
+[answer]结果1[/answer]
+[answer]结果2[/answer]"""
+    
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\]\s*(\d+)\s*\[/answer\]', output)
+        try:
+            return [int(m) % mod for m in matches] if matches else None
+        except:
+            return None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        expected = [q['expected'] for q in identity['queries'] if q['type'] == 2]
+        return solution == expected if solution else False