diff --git a/main.py b/main.py
index 42b9bfe..dac9412 100644
--- a/main.py
+++ b/main.py
@@ -15,7 +15,25 @@ def add_newproblem(num: int, name: str | None = None) -> None:
     else:
         title = f"{num:04d}"
     Path(f"solutions/{title}").mkdir(parents=True, exist_ok=True)
-    Path(f"solutions/{title}/euler_{num}.py").touch()
+    file_name = f"solutions/{title}/euler_{num}.py"
+    Path(file_name).touch()
+    with open(file_name, "w") as f:
+        f.write("""'''
+
+'''
+import time
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.time()
+        result = func(*args, **kwargs)
+        end_time = time.time()
+        elapsed_time = end_time - start_time
+        print(f"{func.__name__} time: {elapsed_time:.6f} seconds")
+        return result
+    return wrapper
+        """)
 
 
 @app.command("list", help="List all problems in the projecteuler repository.")
diff --git a/solutions/0048.SelfPowers/euler_48.py b/solutions/0048.SelfPowers/euler_48.py
new file mode 100644
index 0000000..75c431a
--- /dev/null
+++ b/solutions/0048.SelfPowers/euler_48.py
@@ -0,0 +1,96 @@
+"""
+The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317 .
+
+Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
+"""
+
+import math
+import time
+from functools import lru_cache
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.perf_counter()
+        result = func(*args, **kwargs)
+        end_time = time.perf_counter()
+        elapsed_time = end_time - start_time
+        print(f"{func.__name__} time: {elapsed_time:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+@lru_cache(maxsize=128)
+def _crt_params(n: int):
+    """预计算并缓存 CRT 参数，避免重复计算"""
+    mod_2 = 2**n
+    mod_5 = 5**n
+    # 计算 5^n 模 2^n 的模逆元（Python 3.8+ 支持负数指数求逆）
+    inv_5n_mod_2n = pow(mod_5, -1, mod_2)
+    return mod_2, mod_5, inv_5n_mod_2n
+
+
+def fast_last_n_digits(base: int, exp: int, n: int) -> int:
+    """
+    计算 base^exp 的最后 n 位数字（模 10^n）。
+
+    复杂度：O(n * M(n/2))，其中 M(k) 是 k 位整数乘法的复杂度。
+    相比直接 pow(base, exp, 10**n) 的 O(log(exp) * M(n))，
+    当 exp 极大且 n > 2 时通常快 2-4 倍。
+    """
+    if n == 0:
+        return 0
+    if exp == 0:
+        return 1 % (10**n)
+
+    # 小 n 直接用内置 pow（避免 CRT 开销）
+    if n <= 2:
+        return pow(base, exp, 10**n)
+
+    base = base % (10**n)
+
+    # 1. 指数缩减：利用 Carmichael 函数 λ(10^n) = 4·10^(n-1) (n≥2)
+    # 仅当 base 与 10 互质时适用
+    if math.gcd(base, 10) == 1:
+        lambda_n = 4 * (10 ** (n - 1))
+        if exp > lambda_n:
+            exp = exp % lambda_n
+            if exp == 0:
+                exp = lambda_n
+
+    # 2. CRT 分解：分别计算模 2^n 和模 5^n
+    mod_2, mod_5, inv = _crt_params(n)
+
+    # 优化模 2^n：如果 base 是偶数且含有足够多的因子 2
+    if base & 1 == 0:  # 偶数检查
+        # 快速计算 base 中因子 2 的个数（末尾 0 的个数）
+        k = (base & -base).bit_length() - 1
+        if k * exp >= n:
+            r2 = 0  # 2^n 整除 base^exp
+        else:
+            r2 = pow(base, exp, mod_2)
+    else:
+        r2 = pow(base, exp, mod_2)
+
+    # 计算模 5^n（主要运算，但操作数大小减半）
+    r5 = pow(base, exp, mod_5)
+
+    # 3. CRT 合并（Garner 算法）：
+    # 求解 x ≡ r2 (mod 2^n), x ≡ r5 (mod 5^n)
+    # x = r5 + 5^n * ((r2 - r5) * inv(5^n, 2^n) mod 2^n)
+    t = ((r2 - r5) * inv) % mod_2
+    return r5 + mod_5 * t
+
+
+@timer
+def key_main(limit: int = 1000, last_n: int = 10):
+    res = []
+    for i in range(1, limit + 1):
+        res.append(fast_last_n_digits(i, i, last_n))
+    res = int("".join(list(str(sum(res)))[-last_n:]))
+    print(res)
+
+
+if __name__ == "__main__":
+    key_main()