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internbootcamp/bootcamp/chackit/chackit.py

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+"""# 
+
+### 谜题描述
+Little X has met the following problem recently. 
+
+Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>
+
+Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code: 
+    
+    
+      
+        ans = solve(l, r) % a;  
+        if (ans <= 0)  
+          ans += a;  
+      
+    
+
+This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.
+
+Input
+
+The first line contains a single integer a (1 ≤ a ≤ 1018).
+
+Output
+
+Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
+
+Examples
+
+Input
+
+46
+
+
+Output
+
+1 10
+
+
+Input
+
+126444381000032
+
+
+Output
+
+2333333 2333333333333
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+def f(s):
+    N = 0
+    p = 0
+    L = len(s)
+    for i in xrange(len(s)):
+        k = L - i - 1
+        for j in xrange(int(s[i])):
+            N += 9 * k * (10 ** k) / 2
+            N += (p + j) * (10 ** k)
+        p += int(s[i])
+    return N
+
+def g(N):
+    if N == 0:
+        return '0'
+    s = ''
+    L = 20
+    for i in xrange(L):
+        d = 0
+        for j in xrange(10):
+            if f(s + str(j) + ('0' * (L - i - 1))) >= N:
+                s = s + str(j - 1)
+                d = 1
+                break
+        if not d:
+            s = s + str(9)
+        #print s
+    return s
+
+A = input()
+s = []
+p = []
+i = 1
+while 1:
+    m = g(i * A)
+    q = f(m) % A
+    d = 0
+#    print m, q
+    for i in xrange(len(p)):
+        if q == p[i] and int(m) > int(s[i]):
+            print int(s[i]), int(m) - 1
+            d = 1
+            break
+    if d:
+        break
+    s.append(m)
+    p.append(f(m) % A)
+    i += 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 re
+from bootcamp import Basebootcamp
+
+def f(s):
+    N = 0
+    p = 0
+    L = len(s)
+    for i in range(len(s)):
+        k = L - i - 1
+        for j in range(int(s[i])):
+            term1 = 9 * k * (10 ** k) // 2
+            term2 = (p + j) * (10 ** k)
+            N += term1 + term2
+        p += int(s[i])
+    return N
+
+def g(N):
+    if N == 0:
+        return '0'
+    s = ''
+    L = 200  # 调整为200位
+    for i in range(L):
+        d = 0
+        for j in range(10):
+            test_s = s + str(j) + '0' * (L - i - 1)
+            current_f = f(test_s)
+            if current_f >= N:
+                if j > 0:
+                    s += str(j-1)
+                else:
+                    s += '0'  # 处理j=0的情况
+                d = 1
+                break
+        if not d:
+            s += '9'
+    s = s.lstrip('0') or '0'
+    return s
+
+def find_test_case(a):
+    s_list = []
+    p_list = []
+    i = 1
+    while True:
+        target = i * a
+        m = g(target)
+        q = f(m) % a
+        for idx in range(len(p_list)):
+            if q == p_list[idx] and int(m) > int(s_list[idx]):
+                l = int(s_list[idx])
+                r = int(m) - 1
+                return (l, r)
+        s_list.append(m)
+        p_list.append(q)
+        i += 1
+
+class Chackitbootcamp(Basebootcamp):
+    def __init__(self, max_a=10**18):  # 支持更大的a值
+        self.max_a = max_a
+
+    def case_generator(self):
+        a = random.randint(1, self.max_a)
+        l, r = find_test_case(a)
+        # 确保数值有效性（示例代码逻辑保证）
+        return {'a': a, 'l': l, 'r': r}
+
+    @staticmethod
+    def prompt_func(question_case):
+        a = question_case['a']
+        prompt = f"""Little X需要构造一个hack测试用例。请找到两个整数l和r，使得所有在区间[l, r]内数字的digit sum之和模{a}等于0。
+
+输入要求：
+- 第一行为整数a（此处a={a}）
+- 输出l和r满足：1 ≤ l ≤ r < 10^200
+
+关键规则：
+1. digit sum定义：例如f(123)=1+2+3=6
+2. 目标程序使用错误取模逻辑：ans = solve(l, r) % a; if (ans <=0) ans += a;
+3. 你构造的(l, r)必须使正确结果为0，但目标程序会错误输出a
+
+输出格式：
+将答案放在[answer]标签内，例如：[answer]1 10[/answer]"""
+        return prompt
+
+    @staticmethod
+    def extract_output(output):
+        # 优先提取answer标签内容
+        answer_tag = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
+        if answer_tag:
+            pair = answer_tag[-1].strip().split()
+            if len(pair) == 2:
+                try:
+                    return (int(pair[0]), int(pair[1]))
+                except:
+                    pass
+        
+        # 兜底提取最后的数字对
+        digit_pairs = re.findall(r'\b(\d+)\s+(\d+)\b', output)
+        if digit_pairs:
+            last_pair = digit_pairs[-1]
+            try:
+                return (int(last_pair[0]), int(last_pair[1]))
+            except:
+                return None
+        return None
+
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        if not solution or len(solution) != 2:
+            return False
+        l, r = solution
+        a = identity['a']
+        
+        # 基本约束检查
+        if not (1 <= l <= r < 10**200):
+            return False
+        if any(len(str(x)) != len(str(int(x))) for x in (l, r)):  # 检查前导零
+            return False
+
+        # 优化计算模值
+        def compute_mod(n_str, mod):
+            total = 0
+            pos = 0
+            digit_sum = 0
+            for ch in reversed(n_str):
+                digit = int(ch)
+                cnt = digit
+                # 0-9的模式重复次数
+                full_cycles = cnt * (cnt-1) // 2
+                total += (full_cycles * (10**pos)) % mod
+                # 当前位贡献
+                total += digit_sum * cnt * (10**pos) % mod
+                # 更新digit_sum
+                digit_sum += digit
+                pos += 1
+            return total % mod
+
+        try:
+            mod_total = (compute_mod(str(r), a) - compute_mod(str(l-1), a)) % a
+            return mod_total == 0
+        except:
+            return False