📝 docs(0031)：添加动态规划数学原理和权威文献参考

✨ feat(0032)：新增Pandigital数字问题解决方案

solutions/0032.Pandigital/euler_32.py

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+"""
+We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once;
+for example, the 5-digit number, 15234, is 1 through 5 pandigital.
+
+The product 7254 is unusual, as the identity, 39 × 186 = 7254,
+containing multiplicand, multiplier, and product is 1 through 9 pandigital.
+
+
+Find the sum of all products whose multiplicand/multiplier/product identity can be written
+as a 1 through 9 pandigital.
+HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
+"""
+
+import time
+from itertools import permutations
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start = time.time()
+        result = func(*args, **kwargs)
+        end = time.time()
+        print(f"{func.__name__} took {end - start:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+def is_pandigital(n: list[str]) -> bool:
+    if n[0] == "1":
+        x = [int("".join(n[:2]))]
+        y = [int("".join(n[2:5]))]
+    else:
+        x = [int(n[0]), int("".join(n[:2]))]
+        y = [int("".join(n[1:5])), int("".join(n[2:5]))]
+    z = int("".join(n[5:]))
+    for i in range(len(x)):
+        if x[i] * y[i] == z:
+            return True
+    return False
+
+
+def pandigital_products() -> list[int]:
+    all_permuts = permutations("123456789")
+    res = set()
+    for perm in all_permuts:
+        if is_pandigital(list(perm)):
+            res.add(int("".join(perm[5:])))
+    return list(res)
+
+
+@timer
+def main_compute() -> None:
+    res = set()
+    for x in range(2, 10):
+        for z in range(1234, 9877):
+            if z % x == 0:
+                y = z // x
+                if "".join(sorted(str(x) + str(y) + str(z))) == "123456789":
+                    res.add(z)
+                    print(f"{x} * {y} = {z}")
+    for x in range(12, 99):
+        for z in range(1234, 9877):
+            if z % x == 0:
+                y = z // x
+                if "".join(sorted(str(x) + str(y) + str(z))) == "123456789":
+                    res.add(z)
+                    print(f"{x} * {y} = {z}")
+    res = list(res)
+    print(f"= {sum(res)}")
+
+
+@timer
+def main() -> None:
+    res = pandigital_products()
+    print("+".join(map(str, res)))
+    print(f"= {sum(res)}")
+
+
+if __name__ == "__main__":
+    main()
+    main_compute()

solutions/0032.Pandigital/readme.md

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+比较容易确定这样一点：
+1. 可实现的数位分布：【2，3，4】，【1，4，4】。
+2. 1位数： 2-9
+3. 2位数： 12-98
+4. 4位数： 1234-9876
+
+所以，最简单的就是试算所有可能。我也考虑了使用排列方法进行计算，但是效率比直接计算要略低。