init-commit

internbootcamp/bootcamp/dstring/dstring.py

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+"""# 
+
+### 谜题描述
+One day in the IT lesson Anna and Maria learned about the lexicographic order.
+
+String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages​​.
+
+The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string \"aab\": \"a\", \"a\", \"aa\", \"ab\", \"aab\", \"b\"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
+
+Input
+
+The first line contains a non-empty string that only consists of small Latin letters (\"a\"-\"z\"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
+
+Output
+
+Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying \"No such line.\" (without the quotes).
+
+Examples
+
+Input
+
+aa
+2
+
+
+Output
+
+a
+
+
+Input
+
+abc
+5
+
+
+Output
+
+bc
+
+
+Input
+
+abab
+7
+
+
+Output
+
+b
+
+Note
+
+In the second sample before string \"bc\" follow strings \"a\", \"ab\", \"abc\", \"b\".
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+from heapq import heappush
+from heapq import heappop
+s = raw_input()
+n = input()
+#n, = map(int, raw_input().strip())
+h = []
+ls = len(s)
+if ls * (ls + 1) / 2 < n:
+    print \"No such line.\"
+    exit(0)
+for i in xrange(ls):
+    heappush(h, (s[i], i))
+while n > 0:
+    now = heappop(h)
+    n = n -1
+    if n == 0:
+        print now[0]
+        exit(0)
+    if now[1] < ls - 1:    
+        heappush(h, (now[0] + s[now[1] + 1], now[1] + 1))
+```
+
+
+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
+import heapq
+from bootcamp import Basebootcamp
+
+class Dstringbootcamp(Basebootcamp):
+    def __init__(self, s_len_min=3, s_len_max=10, k_min=1, k_max=100):
+        self.s_len_min = s_len_min
+        self.s_len_max = s_len_max
+        self.k_min = k_min
+        self.k_max = k_max
+
+    def case_generator(self):
+        n = random.randint(self.s_len_min, self.s_len_max)
+        s = ''.join(random.choices(string.ascii_lowercase, k=n))
+        total = n * (n + 1) // 2
+        # Ensure k does not exceed the total number of substrings
+        k = random.randint(self.k_min, min(self.k_max, total))
+        return {
+            's': s,
+            'k': k
+        }
+
+    @staticmethod
+    def prompt_func(question_case):
+        s = question_case['s']
+        k = question_case['k']
+        prompt = f"Anna and Maria are given the string: {s}\n"
+        prompt += f"They need to find the {k}-th lexicographically smallest substring. "
+        prompt += "All possible substrings, including duplicates, should be considered and sorted in lex order. "
+        prompt += "If the total number of substrings is less than {k}, output 'No such line.' "
+        prompt += "Please provide your answer within [answer] and [/answer] tags."
+        return prompt
+
+    @staticmethod
+    def extract_output(output):
+        start = output.rfind('[answer]')
+        if start == -1:
+            return None
+        end = output.find('[/answer]', start)
+        if end == -1:
+            return None
+        return output[start+8:end].strip()
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        s = identity['s']
+        k = identity['k']
+        n = len(s)
+        total = n * (n + 1) // 2
+        if k > total:
+            return solution == "No such line."
+        heap = []
+        for i in range(len(s)):
+            heapq.heappush(heap, (s[i], i))
+        current = None
+        for _ in range(k):
+            current = heapq.heappop(heap)
+            if current[1] < len(s) - 1:
+                next_char = s[current[1] + 1]
+                heapq.heappush(heap, (current[0] + next_char, current[1] + 1))
+        expected = current[0] if k <= total else "No such line."
+        return solution == expected