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internbootcamp/bootcamp/cdiversesubstrings/cdiversesubstrings.py

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+"""# 
+
+### 谜题描述
+String diversity is the number of symbols that occur in the string at least once. Diversity of s will be denoted by d(s). For example , d(\"aaa\")=1, d(\"abacaba\")=3.
+
+Given a string s, consisting of lowercase Latin letters. Consider all its substrings. Obviously, any substring diversity is a number from 1 to d(s). Find statistics about substrings diversity: for each k from 1 to d(s), find how many substrings of s has a diversity of exactly k.
+
+Input
+
+The input consists of a single line containing s. It contains only lowercase Latin letters, the length of s is from 1 to 3·105.
+
+Output
+
+Print to the first line the value d(s). Print sequence t1, t2, ..., td(s) to the following lines, where ti is the number of substrings of s having diversity of exactly i.
+
+Examples
+
+Input
+
+abca
+
+
+Output
+
+3
+4
+3
+3
+
+
+Input
+
+aabacaabbad
+
+
+Output
+
+4
+14
+19
+28
+5
+
+Note
+
+Consider the first example.
+
+We denote by s(i, j) a substring of \"abca\" with the indices in the segment [i, j].
+
+  * s(1, 1) =  \"a\", d(\"a\") = 1
+  * s(2, 2) =  \"b\", d(\"b\") = 1
+  * s(3, 3) =  \"c\", d(\"c\") = 1
+  * s(4, 4) =  \"a\", d(\"a\") = 1
+  * s(1, 2) =  \"ab\", d(\"ab\") = 2
+  * s(2, 3) =  \"bc\", d(\"bc\") = 2
+  * s(3, 4) =  \"ca\", d(\"ca\") = 2
+  * s(1, 3) =  \"abc\", d(\"abc\") = 3
+  * s(2, 4) =  \"bca\", d(\"bca\") = 3
+  * s(1, 4) =  \"abca\", d(\"abca\") = 3
+
+
+
+Total number of substring with diversity 1 is 4, with diversity 2 equals 3, 3 diversity is 3.
+
+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;
+using ll = long long;
+using ull = unsigned long long;
+using vi = vector<int>;
+using vl = vector<long>;
+using vll = vector<ll>;
+using vb = vector<bool>;
+using vvb = vector<vb>;
+using vvi = vector<vector<int> >;
+using ii = pair<int, int>;
+using vii = vector<ii>;
+using vs = vector<string>;
+using msi = map<string, int>;
+using iss = istringstream;
+int main() {
+  ios_base::sync_with_stdio(false);
+  cin.tie(NULL);
+  string s;
+  cin >> s;
+  int n = s.length();
+  for (int i = int(0); i <= int(n - 1); i++) s[i] -= 'a';
+  vb seen(26, false);
+  for (int i = int(0); i <= int(n - 1); i++) seen[s[i]] = true;
+  int ds = accumulate(seen.begin(), seen.end(), 0);
+  vll ans(ds + 1, 0);
+  for (int k = int(1); k <= int(ds); k++) {
+    vector<deque<int> > dq(26, deque<int>());
+    int l = 0, r = -1;
+    int uniq = 0;
+    int upos = 0;
+    ll cnt = 0;
+    while (l < n) {
+      while (true) {
+        if (r == n - 1) break;
+        if (uniq == k && !dq[s[r + 1]].size() > 0) break;
+        r++;
+        if (dq[s[r]].size() == 0) {
+          uniq++;
+          upos = r;
+        }
+        dq[s[r]].push_back(r);
+        if (uniq == k) cnt++;
+      }
+      ans[k] += cnt;
+      dq[s[l]].pop_front();
+      if (dq[s[l]].size() == 0) {
+        uniq--;
+        cnt = 0;
+      } else {
+        int nupos = max(upos, dq[s[l]][0]);
+        cnt = max(0ll, cnt - (nupos - upos));
+        upos = nupos;
+      }
+      l++;
+    }
+  }
+  cout << ds << '\n';
+  for (int k = int(1); k <= int(ds); k++) cout << ans[k] << '\n';
+  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 string
+from collections import defaultdict, deque
+from bootcamp import Basebootcamp
+
+class Cdiversesubstringsbootcamp(Basebootcamp):
+    def __init__(self, min_length=4, max_length=100):
+        self.min_length = max(min_length, 1)
+        self.max_length = max(max_length, self.min_length)
+
+    def case_generator(self):
+        # Generate string with controlled diversity
+        length = random.randint(self.min_length, self.max_length)
+        max_possible_d = min(26, length)
+        target_d = random.randint(1, max_possible_d)
+        
+        # Ensure exactly target_d distinct characters
+        base_chars = random.sample(string.ascii_lowercase, target_d)
+        s_list = base_chars.copy()
+        
+        # Fill remaining length with random choices from base chars
+        for _ in range(length - target_d):
+            s_list.append(random.choice(base_chars))
+        
+        random.shuffle(s_list)
+        s = ''.join(s_list)
+        
+        # Calculate actual d and t_list using optimized algorithm
+        d = len(set(s))
+        t_list = self.compute_t_list(s)
+        
+        return {
+            's': s,
+            'd': d,
+            't_list': t_list
+        }
+
+    def compute_t_list(self, s):
+        n = len(s)
+        s = [ord(c) - ord('a') for c in s]
+        total_d = len(set(s))
+        ans = [0] * (total_d + 1)  # ans[0] unused
+        
+        for k in range(1, total_d + 1):
+            count = defaultdict(int)
+            distinct = 0
+            left = 0
+            res = 0
+            
+            for right in range(n):
+                c = s[right]
+                if count[c] == 0:
+                    distinct += 1
+                count[c] += 1
+                
+                while distinct > k:
+                    left_c = s[left]
+                    count[left_c] -= 1
+                    if count[left_c] == 0:
+                        distinct -= 1
+                    left += 1
+                
+                res += right - left + 1
+            
+            prev_res = 0
+            if k > 1:
+                prev_res = self.at_most_k(s, k-1)
+            ans[k] = res - prev_res
+        
+        return [ans[k] for k in range(1, total_d+1)]
+
+    def at_most_k(self, s, k):
+        count = defaultdict(int)
+        distinct = 0
+        left = 0
+        res = 0
+        
+        for right in range(len(s)):
+            c = s[right]
+            if count[c] == 0:
+                distinct += 1
+            count[c] += 1
+            
+            while distinct > k:
+                left_c = s[left]
+                count[left_c] -= 1
+                if count[left_c] == 0:
+                    distinct -= 1
+                left += 1
+            
+            res += right - left + 1
+        
+        return res
+
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        s = question_case['s']
+        return f"""You are given a string consisting of lowercase Latin letters. The diversity of a string is defined as the number of distinct characters it contains. Your task is to calculate two things: 
+1. The diversity value d(s) of the given string.
+2. For each k from 1 to d(s), determine how many substrings have exactly k distinct characters.
+
+Input string: {s}
+
+Output Format:
+- First line: d(s)
+- Next d(s) lines: Each line contains the count of substrings with diversity exactly k (k from 1 to d(s))
+
+Enclose your final answer within [answer] and [/answer] tags. For example:
+[answer]
+3
+4
+3
+3
+[/answer]"""
+
+    @staticmethod
+    def extract_output(output):
+        import re
+        matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
+        if not matches:
+            return None
+        
+        numbers = []
+        last_answer = matches[-1].strip()
+        for line in last_answer.split('\n'):
+            line = line.strip()
+            if line:
+                try:
+                    numbers.append(int(line))
+                except ValueError:
+                    return None
+        
+        return numbers if numbers else None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        if not isinstance(solution, list) or len(solution) != identity['d'] + 1:
+            return False
+        if solution[0] != identity['d']:
+            return False
+        return solution[1:] == identity['t_list']