init-commit

internbootcamp/bootcamp/cdoublehappiness/cdoublehappiness.py

@@ -0,0 +1,161 @@
+"""# 
+
+### 谜题描述
+On the math lesson a teacher asked each pupil to come up with his own lucky numbers. As a fan of number theory Peter chose prime numbers. Bob was more original. He said that number t is his lucky number, if it can be represented as: 
+
+t = a2 + b2,  where a, b are arbitrary positive integers.
+
+Now, the boys decided to find out how many days of the interval [l, r] (l ≤ r) are suitable for pair programming. They decided that the day i (l ≤ i ≤ r) is suitable for pair programming if and only if the number i is lucky for Peter and lucky for Bob at the same time. Help the boys to find the number of such days.
+
+Input
+
+The first line of the input contains integer numbers l, r (1 ≤ l, r ≤ 3·108).
+
+Output
+
+In the only line print the number of days on the segment [l, r], which are lucky for Peter and Bob at the same time.
+
+Examples
+
+Input
+
+3 5
+
+
+Output
+
+1
+
+
+Input
+
+6 66
+
+
+Output
+
+7
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+f = lambda x: int(x**.5)+1
+h,z = map(bytearray,'1\0')
+
+L,R = map(int,raw_input().split())
+
+n = f(R); b = n/2*h; b[0] = 0
+for k in xrange(1,f(n)/2):
+    if b[k]: p = 2*k+1; s = k*(p+1); b[s::p] = z*len(b[s::p])
+
+g = ((i*(i+1),2*i+1) for i,v in enumerate(b) if v)
+
+r = (R+3)/4*h; r[0] = 0
+for s,p in g:
+    r[s::p] = z*len(r[s::p])
+
+print r.count(h,(L+2)/4)+(L<=2<=R)
+```
+
+
+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 math import isqrt
+from bootcamp import Basebootcamp
+
+class Cdoublehappinessbootcamp(Basebootcamp):
+    def __init__(self, max_r=10**5):
+        """
+        初始化训练场参数，默认最大右边界为1e5以保证生成效率
+        :param max_r: 生成区间右端点最大值，默认为100000
+        """
+        self.max_r = max_r
+    
+    def _sieve(self, l, r):
+        """高效生成区间[l, r]内的质数列表"""
+        if r < 2:
+            return []
+        
+        # 生成基础质数用于筛法
+        limit = isqrt(r) + 1
+        sieve = [True] * (limit + 1)
+        sieve[0:2] = [False, False]
+        for i in range(2, isqrt(limit) + 1):
+            if sieve[i]:
+                sieve[i*i : limit+1 : i] = [False] * len(sieve[i*i : limit+1 : i])
+        base_primes = [i for i, prime in enumerate(sieve) if prime]
+        
+        # 区间筛法
+        segment_size = r - l + 1
+        sieve = [True] * segment_size
+        for p in base_primes:
+            start = max(p * p, ((l + p - 1) // p) * p)
+            for i in range(start, r+1, p):
+                sieve[i - l] = False
+        
+        # 处理小质数的平方
+        for i in range(max(2, l), isqrt(r) + 1):
+            if sieve[i - l]:
+                for j in range(i*i, r+1, i):
+                    sieve[j - l] = False
+        
+        return [i + l for i in range(segment_size) if sieve[i]]
+    
+    def case_generator(self):
+        """生成随机区间案例并准确计算答案"""
+        while True:
+            # 生成有效区间
+            l = random.randint(1, self.max_r - 1)
+            r = random.randint(l, self.max_r)
+            
+            # 获取区间质数
+            primes = self._sieve(l, r)
+            
+            # 计算有效质数数量
+            valid_primes = [p for p in primes if p == 2 or p % 4 == 1]
+            
+            # 确保生成有效案例（可包含0个解）
+            return {
+                'l': l,
+                'r': r,
+                'answer': len(valid_primes)
+            }
+    
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        l = question_case['l']
+        r = question_case['r']
+        return f"""你是数学课代表，需要帮助Peter和Bob计算在区间[{l}, {r}]内同时满足两个条件的数字：
+1. Peter条件：必须是质数（只能被1和自身整除）
+2. Bob条件：可以表示为两个正整数的平方和（即存在a,b>0使得t=a²+b²）
+
+重要数学规则：
+- 根据费马定理，质数p可表示为两个平方数之和当且仅当p=2或p≡1(mod4)
+
+请逐步分析后输出准确答案，并将最终答案放在[answer]和[/answer]标签之间。
+
+例如：
+当输入为6 66时，正确质数是13,17,29,37,41,53,61，因此输出为7
+
+当前需要解决的输入：
+l = {l}, r = {r}
+
+请按照以下格式输出：
+[answer]答案数字[/answer]"""
+    
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\]\s*(\d+)\s*\[/answer\]', output, re.IGNORECASE)
+        if matches:
+            try:
+                return int(matches[-1])
+            except ValueError:
+                return None
+        return None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        return solution == identity['answer']