diff --git a/solutions/0043.SubStrDivisibility/euler_43.py b/solutions/0043.SubStrDivisibility/euler_43.py
new file mode 100644
index 0000000..fe2df44
--- /dev/null
+++ b/solutions/0043.SubStrDivisibility/euler_43.py
@@ -0,0 +1,110 @@
+"""
+The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9
+in some order, but it also has a rather interesting sub-string divisibility property.
+
+Let d_1 be the 1st digit, d_2 be the 2nd digit, and so on. In this way, we note the following:
+
+    d_2d_3d_4 is divisible by 2
+    d_3d_4d_5 is divisible by 3
+    d_4d_5d_6 is divisible by 5
+    d_5d_6d_7 is divisible by 7
+    d_6d_7d_8 is divisible by 11
+    d_7d_8d_9 is divisible by 13
+    d_8d_9d_10 is divisible by 17
+
+Find the sum of all 0 to 9 pandigital numbers with this property.
+"""
+
+import time
+from itertools import permutations
+from tkinter.constants import TOP
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.perf_counter()
+        result = func(*args, **kwargs)
+        end_time = time.perf_counter()
+        print(f"Execution time: {end_time - start_time} seconds")
+        return result
+
+    return wrapper
+
+
+def get_all_numbers() -> list[int]:
+    ori = list(permutations(list(range(10))))
+    res = []
+    for x in ori:
+        if x[3] % 2 != 0:
+            continue
+        if x[5] not in [0, 5]:
+            continue
+        for i, d in enumerate([2, 3, 5, 7, 11, 13, 17]):
+            if int("".join(map(str, x[i + 1 : i + 4]))) % d != 0:
+                break
+        else:
+            res.append(int("".join(map(str, x))))
+    return res
+
+
+def optimized_solve() -> tuple[int, list[int]]:
+    total = 0
+
+    # 从最后三位开始，这些必须能被17整除
+    candidates = []
+    for n in range(102, 1000, 17):
+        s = str(n).zfill(3)
+        if len(set(s)) == 3:  # 三位数字必须不同
+            candidates.append(s)
+
+    # 逐步向前添加数字，检查约束
+    for i, prime in enumerate([13, 11, 7, 5, 3, 2]):
+        new_candidates = []
+        for cand in candidates:
+            # 获取当前已使用的数字
+            used_digits = set(cand)
+
+            # 尝试添加一个新数字在最前面
+            for d in "0123456789":
+                if d not in used_digits:
+                    # 新形成的三位数
+                    new_triplet = d + cand[:2]
+                    # 检查是否能被对应的质数整除
+                    if int(new_triplet) % prime == 0:
+                        new_candidates.append(d + cand)
+
+        candidates = new_candidates
+
+    # 现在candidates中包含的是d_2到d_10（9位数字）
+    # 我们需要添加第一个数字d_1
+    final_numbers = []
+    for cand in candidates:
+        used_digits = set(cand)
+        # 找出缺失的那个数字
+        for d in "0123456789":
+            if d not in used_digits and d != "0":  # d_1不能为0
+                num = d + cand
+                # 验证这是一个0-9的pandigital数
+                if len(set(num)) == 10:
+                    final_numbers.append(num)
+
+    # 计算总和
+    total = sum(int(num) for num in final_numbers)
+    return total, [int(num) for num in final_numbers]
+
+
+@timer
+def main():
+    numbers = get_all_numbers()
+    print(sum(numbers))
+
+
+@timer
+def main_op():
+    total, _ = optimized_solve()
+    print(total)
+
+
+if __name__ == "__main__":
+    main()
+    main_op()
diff --git a/solutions/0043.SubStrDivisibility/readme.md b/solutions/0043.SubStrDivisibility/readme.md
new file mode 100644
index 0000000..b69dcd6
--- /dev/null
+++ b/solutions/0043.SubStrDivisibility/readme.md
@@ -0,0 +1,8 @@
+# 简单思路说明
+
+使用简单方式循环构建进行计算，刨除最简单的特征外，最多减少一半计算时间。这是从前往后进行计算的时候。
+
+但是从后往前，就要快速多了。最后三位是形式确认的17的倍数，所以可以减少更多的计算次数，
+（一千以内的17倍数只有不足60个），相应的递归到下一位数，所需计算次数也不会慢更多。
+
+从最小可确认开始思考这种具有递归特征的问题，可能更容易进行计算求解。