diff --git a/solutions/0024.Permutations/README.md b/solutions/0024.Permutations/README.md
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+++ b/solutions/0024.Permutations/README.md
@@ -0,0 +1,14 @@
+# 有序排列
+
+我没想到的是，排列也有很多事情是我没想到的，比如 [欧拉数 Eulerian Number](https://en.wikipedia.org/wiki/Eulerian_number)。
+在 [OEIS A008292](https://oeis.org/A008292) 还有更多讨论。
+
+
+本来以为使用python自带的排列方法，就是最快的了。但没想到欧拉数的有序计算逻辑似乎更好用：
+
+对于 $n$ 个元素的排列：
+- 总共有 $n!$ 种排列
+- 以某个特定元素开头的排列有 $(n−1)!$ 种
+- 这构成了变进制的基础，可以用来计算第 $k$ 个排列。
+
+相比在序列上移动要快速多了，毕竟是与原序列的长度相关，而不与位置相关。
diff --git a/solutions/0024.Permutations/euler_24.py b/solutions/0024.Permutations/euler_24.py
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+++ b/solutions/0024.Permutations/euler_24.py
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+"""
+A permutation is an ordered arrangement of objects.
+For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4.
+If all of the permutations are listed numerically or alphabetically, we call it lexicographic order.
+The lexicographic permutations of 0, 1 and 2 are:
+012   021   102   120   201   210
+
+What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
+"""
+
+import time
+from itertools import permutations
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.time()
+        result = func(*args, **kwargs)
+        end_time = time.time()
+        print(f"Execution time: {end_time - start_time:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+def ordered_permutations(n: int) -> list[int]:
+    digits = list(range(10))
+    all_permutations = permutations(digits)
+    for _ in range(n - 1):
+        next(all_permutations)
+    return next(all_permutations)
+
+
+@timer
+def main_iter():
+    n = 1000000
+    result = ordered_permutations(n)
+    print(f"used iter: {''.join(map(str, result))}")
+
+
+def get_permutation(seq, k):
+    """
+    返回序列 seq 的第 k 个字典序排列 (1-based)
+
+    参数:
+        seq: 可迭代对象（列表、元组、字符串等），元素需有序
+        k: 第 k 个排列（从1开始计数）
+
+    返回:
+        与输入类型相同的排列结果
+    """
+    n = len(seq)
+
+    # 预计算阶乘
+    factorial = [1]
+    for i in range(1, n):
+        factorial.append(factorial[-1] * i)
+
+    # 将输入转为列表（复制一份避免修改原序列）
+    elements = list(seq)
+    k -= 1  # 转为0-based索引
+
+    # 构建结果
+    result = []
+    for i in range(n, 0, -1):
+        index = k // factorial[i - 1]  # 当前位选择的元素索引
+        result.append(elements.pop(index))
+        k %= factorial[i - 1]  # 更新剩余位
+
+    # 根据输入类型返回相应格式
+    if isinstance(seq, str):
+        return "".join(result)
+    return result
+
+
+@timer
+def main_math():
+    n = 1000000
+    result = get_permutation("0123456789", n)
+    print(f"used math: {result}")
+
+
+if __name__ == "__main__":
+    main_iter()
+    main_math()
diff --git a/solutions/0024/euler_24.py b/solutions/0024/euler_24.py
deleted file mode 100644
index 02640d0..0000000
--- a/solutions/0024/euler_24.py
+++ /dev/null
@@ -1,29 +0,0 @@
-"""
-A permutation is an ordered arrangement of objects.
-For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4.
-If all of the permutations are listed numerically or alphabetically, we call it lexicographic order.
-The lexicographic permutations of 0, 1 and 2 are:
-012   021   102   120   201   210
-
-What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
-"""
-
-from itertools import permutations
-
-
-def ordered_permutations(n: int) -> list[int]:
-    digits = list(range(10))
-    all_permutations = permutations(digits)
-    for _ in range(n - 1):
-        next(all_permutations)
-    return next(all_permutations)
-
-
-def main():
-    n = 1000000
-    result = ordered_permutations(n)
-    print("".join(map(str, result)))
-
-
-if __name__ == "__main__":
-    main()
diff --git a/solutions/0025.Fibonacci/euler_25.py b/solutions/0025.Fibonacci/euler_25.py
new file mode 100644
index 0000000..01a0b5f
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+++ b/solutions/0025.Fibonacci/euler_25.py
@@ -0,0 +1,77 @@
+"""
+The Fibonacci sequence is defined by the recurrence relation:
+F_n = F_{n-1} + F_{n-2}, where F_1 = 1 and F_2 = 1.
+
+Hence the first 12 terms will be: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144
+The 12th term, F_12, is the first term to contain three digits.
+
+What is the index of the first term in the Fibonacci sequence to contain 1000-digits?
+"""
+
+import math
+import time
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.time()
+        result = func(*args, **kwargs)
+        end_time = time.time()
+        during = end_time - start_time
+        if during > 1:
+            print(f"function {func.__name__} taken: {during:.6f} seconds")
+        elif during > 0.001:
+            print(f"function {func.__name__} taken: {during * 1000:.3f} milliseconds")
+        else:
+            print(
+                f"function {func.__name__} taken: {during * 1000000:.3f} microseconds"
+            )
+        return result
+
+    return wrapper
+
+
+def big_sum(a: str, b: str) -> str:
+    length = max(len(a), len(b))
+    carry = 0
+    result = []
+    for i in range(length):
+        digit_a = int(a[-i - 1]) if i < len(a) else 0
+        digit_b = int(b[-i - 1]) if i < len(b) else 0
+        sum_digit = digit_a + digit_b + carry
+        result.append(str(sum_digit % 10))
+        carry = sum_digit // 10
+    if carry:
+        result.append(str(carry))
+    return "".join(result[::-1])
+
+
+def find_fibonacci_index(digits: int) -> int:
+    a, b = "1", "1"
+    index = 1
+    while len(a) < digits:
+        a, b = b, big_sum(a, b)
+        index += 1
+    return index
+
+
+@timer
+def main_in_bigint():
+    print(find_fibonacci_index(1000))
+
+
+def binet(n: int) -> int:
+    index = math.ceil(
+        (math.log10(math.sqrt(5)) + (n - 1)) / math.log10(0.5 * (1 + math.sqrt(5)))
+    )
+    return index
+
+
+@timer
+def main_use_binet():
+    print(f"{binet(1000)}")
+
+
+if __name__ == "__main__":
+    main_in_bigint()
+    main_use_binet()