feat(cli): 支持多重目录结构与特殊解法运行，新增解法列表选项

solutions/0049.PrimePermutations/euler_49.py

@@ -70,8 +70,8 @@ def main():
     for i, v in enumerate(primes):
         if v == max_sp:
             break
-        gap = int((max_sp - v) / 3)
-        for g in range(gap, (primes[i + 1] - v - 1), -1):
+        gap = (max_sp - v) // 2
+        for g in range(gap, 0, -1):
             if v + g in primes and v + 2 * g in primes:
                 if list_com(list(str(v)), list(str(v + g))):
                     if list_com(list(str(v)), list(str(v + 2 * g))):

solutions/0049.PrimePermutations/euler_49_better.py

@@ -0,0 +1,106 @@
+"""
+The arithmetic sequence, 1487,4817,8147, in which each of the terms increases by 3330,
+is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the
+4-digit numbers are permutations of one another.
+
+There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property,
+but there is one other 4-digit increasing sequence.
+
+What 12-digit number do you form by concatenating the three terms in this sequence?
+"""
+
+import time
+from logging import root
+from pathlib import Path
+
+
+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_ultra_optimized(n: int = 4) -> list[str]:
+    limit = 10**n
+    mini_limit = 10 ** (n - 1) + 1
+    # ========== 1. Bitset 筛法（内存高效） ==========
+    is_prime = bytearray(b"\x01") * limit
+    is_prime[0] = is_prime[1] = 0
+
+    # 埃氏筛，使用切片赋值加速
+    for i in range(2, int(limit**0.5) + 1):
+        if is_prime[i]:
+            start = i * i
+            is_prime[start:limit:i] = b"\x00" * ((limit - start - 1) // i + 1)
+
+    # ========== 2. 按数字签名分组 ==========
+    groups = {}
+    for p in range(mini_limit, limit, 2):  # 只遍历奇数，4位质数必为奇数
+        if is_prime[p]:
+            key = "".join(sorted(str(p)))
+            groups.setdefault(key, []).append(p)
+
+    # ========== 3. 组内搜索（利用模18剪枝） ==========
+    res = []
+    for key, group in groups.items():
+        if len(group) < 3:
+            continue
+
+        group.sort()
+        group_set = set(group)  # 用于O(1)成员检查
+
+        n = len(group)
+        for i in range(n):
+            a = group[i]
+            for j in range(i + 1, n):
+                b = group[j]
+                d = b - a
+
+                # 核心数学优化：公差必须是18的倍数
+                # 原因：(1) 排列数字和相同 => 模9同余 => d%9==0
+                #      (2) 质数>2都是奇数 => 奇+偶=奇 => d%2==0
+                if d % 18 != 0:
+                    continue
+
+                c = b + d
+
+                # 边界检查（利用group有序性提前终止）
+                if c >= limit:
+                    break
+
+                # Bitset O(1) 质数检查
+                if not is_prime[c]:
+                    continue
+
+                # 确认第三项在组内（数字签名匹配）
+                if c in group_set:
+                    res.append(f"{a}-{b}-{c}:{d}")
+
+    return res
+
+
+if __name__ == "__main__":
+    try:
+        n = int(input("Search digits (default is 4-digit): ") or "4")
+    except ValueError:
+        n = 4
+    res = solve_ultra_optimized(n)
+    if lr := len(res) > 10:
+        head = 10
+        print(
+            f"Too much data, only showing the first ten rows. Other data saved in result_{n}_digit.txt."
+        )
+        root_path = Path(__file__).parent
+        with open(f"{root_path}/result_{n}_digit.txt", "w", encoding="utf-8") as f:
+            f.write("\n".join(res))
+    else:
+        head = lr
+    for x in res[:head]:
+        print(x)