diff --git a/solutions/0788.dominatingNambers/README.md b/solutions/0788.dominatingNambers/README.md
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+核心思想是n位数中，n/2个数是相同的数量。
+这就可以转化为一个组合的问题。由此延展成为解法。
diff --git a/solutions/0788.dominatingNambers/euler_788.py b/solutions/0788.dominatingNambers/euler_788.py
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+"""
+A dominating number is a positive integer that has more than half of its digits equal.
+
+For example, 2022is a dominating number because three of its four digits are equal to 2. But
+2021 is not a dominating number.
+
+Let D(N) be how many dominating numbers are less than 10^N .
+For example, D(4)=603 and D(10)=21893256 .
+
+Find D(2022). Give your answer modulo 1_000_000_007.
+"""
+
+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
+
+
+@timer
+def solve(N: int = 2022, MOD: int = 10**9 + 7):
+    # 预处理阶乘和逆阶乘到 N+1
+    fact = [1] * (N + 2)
+    for i in range(1, N + 2):
+        fact[i] = fact[i - 1] * i % MOD
+
+    inv_fact = [1] * (N + 2)
+    inv_fact[N + 1] = pow(fact[N + 1], MOD - 2, MOD)
+    for i in range(N, -1, -1):
+        inv_fact[i] = inv_fact[i + 1] * (i + 1) % MOD
+
+    # 组合数函数
+    def C(n: int, k: int) -> int:
+        if k < 0 or k > n:
+            return 0
+        return fact[n] * inv_fact[k] % MOD * inv_fact[n - k] % MOD
+
+    # 预处理 9 的幂（到 N+2 足够）
+    pow9 = [1] * (N + 3)
+    for i in range(1, N + 3):
+        pow9[i] = pow9[i - 1] * 9 % MOD
+
+    ans = 0
+    K_max = (N - 1) // 2
+    for k in range(0, K_max + 1):
+        term = pow9[k + 1] * (C(N + 1, k + 1) - C(2 * k + 1, k + 1)) % MOD
+        ans = (ans + term) % MOD
+    return ans
+
+
+def main():
+    n = int(input("N:"))
+    mod = int(input("mod:") or (10**9 + 7))
+    print(solve(n, mod))
+
+
+if __name__ == "__main__":
+    main()