✨ 新增欧拉第33题“数字约分分数”双解法实现

solutions/0033.DigitCancelling/euler_33.py

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+"""
+The fraction 49/98 is a curious fraction,
+as an inexperienced mathematician in attempting to simplify it may incorrectly believe
+that 49/98 = 4/8 , which is correct, is obtained by cancelling the 9s.
+
+We shall consider fractions like, 30/50=3/5 , to be trivial examples.
+
+There are exactly four non-trivial examples of this type of fraction,
+less than one in value, and containing two digits in the numerator and denominator.
+
+If the product of these four fractions is given in its lowest common terms,
+find the value of the denominator.
+"""
+
+import time
+from functools import reduce
+from math import gcd
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.time()
+        result = func(*args, **kwargs)
+        end_time = time.time()
+        print(f"{func.__name__} taken: {end_time - start_time:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+@timer
+def main() -> None:
+    res = list()
+    for numerator in range(12, 100):
+        for denominator in range(numerator * 2, 100):
+            if numerator % 10 == 0 and denominator % 10 == 0:
+                continue
+            all_nums = set(
+                [numerator // 10, numerator % 10, denominator // 10, denominator % 10]
+            )
+            if len(all_nums) == 3:
+                a = [i for i in str(numerator) if i not in str(denominator)][0]
+                b = [i for i in str(denominator) if i not in str(numerator)][0]
+                if a != "0" and b != "0":
+                    if (int(a) / int(b)) == (numerator / denominator):
+                        print(f"{numerator}/{denominator}")
+                        res.append((numerator, denominator))
+    res = reduce(lambda x, y: (x[0] * y[0], x[1] * y[1]), res)
+    res = res[1] // gcd(res[0], res[1])
+    print(f"Answer: {res}")
+
+
+@timer
+def main_two() -> None:
+    results = []
+
+    for numerator in range(11, 100):
+        for denominator in range(numerator + 1, 100):  # 分母必须大于分子
+            # 检查是否构成非平凡的情况（有共同数字但不是都为0）
+            num_str, den_str = str(numerator), str(denominator)
+            common_digits = set(num_str) & set(den_str)
+
+            # 排除平凡情况和没有共同数字的情况
+            if len(common_digits) != 1 or "0" in common_digits:
+                continue
+
+            common_digit = list(common_digits)[0]
+
+            # 构造约分后的分数
+            simplified_num = int(num_str.replace(common_digit, "", 1))
+            simplified_den = int(den_str.replace(common_digit, "", 1))
+
+            # 避免除以0
+            if simplified_den == 0:
+                continue
+
+            # 检查分数是否相等
+            if numerator * simplified_den == denominator * simplified_num:
+                print(f"{numerator}/{denominator} = {simplified_num}/{simplified_den}")
+                results.append((numerator, denominator))
+
+    # 计算最终结果（所有分数乘积的分母）
+    product_num = reduce(lambda x, y: x * y[0], results, 1)
+    product_den = reduce(lambda x, y: x * y[1], results, 1)
+
+    # 化简分数
+    common_divisor = gcd(product_num, product_den)
+    final_answer = product_den // common_divisor
+    print(f"Answer: {final_answer}")
+
+
+if __name__ == "__main__":
+    main()
+    main_two()