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internbootcamp/bootcamp/emikeandfoam/emikeandfoam.py

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+"""# 
+
+### 谜题描述
+Mike is a bartender at Rico's bar. At Rico's, they put beer glasses in a special shelf. There are n kinds of beer at Rico's numbered from 1 to n. i-th kind of beer has ai milliliters of foam on it.
+
+<image>
+
+Maxim is Mike's boss. Today he told Mike to perform q queries. Initially the shelf is empty. In each request, Maxim gives him a number x. If beer number x is already in the shelf, then Mike should remove it from the shelf, otherwise he should put it in the shelf.
+
+After each query, Mike should tell him the score of the shelf. Bears are geeks. So they think that the score of a shelf is the number of pairs (i, j) of glasses in the shelf such that i < j and <image> where <image> is the greatest common divisor of numbers a and b.
+
+Mike is tired. So he asked you to help him in performing these requests.
+
+Input
+
+The first line of input contains numbers n and q (1 ≤ n, q ≤ 2 × 105), the number of different kinds of beer and number of queries.
+
+The next line contains n space separated integers, a1, a2, ... , an (1 ≤ ai ≤ 5 × 105), the height of foam in top of each kind of beer.
+
+The next q lines contain the queries. Each query consists of a single integer integer x (1 ≤ x ≤ n), the index of a beer that should be added or removed from the shelf.
+
+Output
+
+For each query, print the answer for that query in one line.
+
+Examples
+
+Input
+
+5 6
+1 2 3 4 6
+1
+2
+3
+4
+5
+1
+
+
+Output
+
+0
+1
+3
+5
+6
+2
+
+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;
+static const int maxn = 5e5 + 6;
+int sp[maxn];
+vector<int> divisor[maxn];
+void seive() {
+  for (int i = 2; i < maxn; i++) sp[i] = i;
+  for (int i = 2; i < maxn; i++) {
+    for (int j = 1; 1LL * i * j < maxn; j++) {
+      sp[i * j] = min(sp[i * j], sp[i]);
+      divisor[i * j].push_back(i);
+    }
+  }
+}
+signed main() {
+  ios_base::sync_with_stdio(false);
+  cin.tie(nullptr);
+  seive();
+  int n, q;
+  cin >> n >> q;
+  vector<int> arr(n + 1);
+  for (int i = 1; i <= n; i++) cin >> arr[i];
+  vector<int> in_self(n + 1);
+  int in_self_cnt = 0;
+  vector<int> divi(maxn);
+  auto get = [&](vector<int> &vec) {
+    int n = vec.size();
+    long long coprime = 0;
+    for (int mask = 0; mask < (1 << n); mask++) {
+      long long d = 1;
+      int bitCnt = 0;
+      for (int i = 0; i < n; i++) {
+        if ((mask >> i) & 1) {
+          d *= vec[i];
+          bitCnt++;
+        }
+      }
+      if (bitCnt & 1)
+        coprime += divi[d];
+      else
+        coprime -= divi[d];
+    }
+    return coprime;
+  };
+  long long ans = 0;
+  while (q--) {
+    int pos;
+    cin >> pos;
+    vector<int> prime_factors;
+    int num = arr[pos];
+    while (num > 1) {
+      int x = sp[num];
+      prime_factors.push_back(x);
+      while (num % x == 0) num /= x;
+    }
+    sort(prime_factors.begin(), prime_factors.end());
+    num = arr[pos];
+    if (in_self[pos]) {
+      for (int d : divisor[num]) divi[d]--;
+      in_self[pos] = 0;
+      in_self_cnt--;
+      ans -= in_self_cnt - get(prime_factors);
+    } else {
+      ans += in_self_cnt - get(prime_factors);
+      for (int d : divisor[num]) divi[d]++;
+      in_self[pos] = 1;
+      in_self_cnt++;
+    }
+    cout << ans << endl;
+  }
+}
+```
+
+
+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 math
+import random
+from collections import defaultdict
+from bootcamp import Basebootcamp
+
+class Emikeandfoambootcamp(Basebootcamp):
+    def __init__(self, max_n=5, max_q=6, max_ai=20):
+        self.max_n = max_n
+        self.max_q = max_q
+        self.max_ai = max_ai
+
+    def case_generator(self):
+        n = random.randint(1, self.max_n)
+        q = random.randint(1, self.max_q)
+        a = [random.randint(1, self.max_ai) for _ in range(n)]
+        queries = [random.randint(1, n) for _ in range(q)]
+        return {'n': n, 'q': q, 'a': a, 'queries': queries}
+
+    @staticmethod
+    def prompt_func(case) -> str:
+        return f"""Mike需要管理啤酒货架。现有{case['n']}种啤酒（编号1-{case['n']}），泡沫量分别为：{' '.join(map(str, case['a']))}。
+处理{case['q']}次查询（数字表示切换对应啤酒的状态），每次查询后计算货架中满足i<j且gcd(a_i,a_j)=1的啤酒对数。
+
+输入格式：
+1行：{case['n']} {case['q']}
+2行：{' '.join(map(str, case['a']))}
+后续{case['q']}行：{' '.join(map(str, case['queries']))}
+
+将每次查询后的结果依次写在[answer]和[/answer]之间，每个结果占一行。示例：
+[answer]
+0
+2
+5
+[/answer]"""
+
+    @staticmethod
+    def extract_output(output):
+        import re
+        matches = re.findall(r'\[answer\](.*?)\[\/answer\]', output, re.DOTALL)
+        if not matches:
+            return None
+        answers = []
+        for line in matches[-1].strip().split('\n'):
+            if line.strip().isdigit():
+                answers.append(int(line.strip()))
+        return answers if answers else None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        expected = cls._compute_expected(
+            identity['n'], identity['q'], 
+            identity['a'], identity['queries']
+        )
+        return solution == expected
+
+    @classmethod
+    def _compute_expected(cls, n, q, a, queries):
+        # 预计算每个数的质因数分解
+        prime_factors_list = []
+        max_ai = max(a) if a else 1
+        sieve = cls._build_sieve(max_ai)
+
+        for num in a:
+            factors = set()
+            temp = num
+            while temp > 1:
+                p = sieve[temp]
+                factors.add(p)
+                while temp % p == 0:
+                    temp //= p
+            prime_factors_list.append(sorted(factors))
+
+        # 初始化状态
+        in_self = defaultdict(bool)
+        divi_counts = defaultdict(int)
+        current_total = 0
+        answer = 0
+        output = []
+
+        for x in queries:
+            idx = x-1  # queries是1-based
+            num = a[idx]
+            factors = prime_factors_list[idx]
+
+            if in_self[idx]:
+                # 移除操作
+                sign = -1
+                in_self[idx] = False
+            else:
+                # 添加操作
+                sign = +1
+                in_self[idx] = True
+
+            # 计算当前贡献
+            coprime_count = 0
+            k = len(factors)
+            for mask in range(1, 1 << k):
+                d = 1
+                bits = 0
+                for i in range(k):
+                    if mask & (1 << i):
+                        d *= factors[i]
+                        bits += 1
+                cnt = divi_counts[d]
+                coprime_count += cnt if bits % 2 else -cnt
+
+            delta = sign * (current_total - coprime_count)
+            answer += delta
+            output.append(answer)
+
+            # 更新除数计数
+            for mask in cls._generate_divisors(num):
+                divi_counts[mask] += sign
+
+            current_total += sign
+
+        return output
+
+    @staticmethod
+    def _build_sieve(max_num):
+        sieve = list(range(max_num+1))
+        for i in range(2, int(math.sqrt(max_num))+1):
+            if sieve[i] == i:
+                for j in range(i*i, max_num+1, i):
+                    if sieve[j] == j:
+                        sieve[j] = i
+        return sieve
+
+    @staticmethod
+    def _generate_divisors(num):
+        if num == 1:
+            return []
+        divisors = set()
+        for i in range(2, int(math.isqrt(num)) + 1):
+            if num % i == 0:
+                divisors.update({i, num//i})
+        divisors.add(num)
+        return divisors