✨ feat(Palindromes)：添加欧拉项目第36题解决方案

📝 docs(Palindromes)：创建README文档说明性能对比

solutions/0036.Palindromes/euler_36.hs

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+{-# LANGUAGE BangPatterns #-}
+
+import Data.Time (diffUTCTime, getCurrentTime)
+import Text.Printf (printf)
+import Data.Bits (shiftR, testBit, setBit)
+import Control.DeepSeq (force, NFData)
+import Control.Exception (evaluate)
+
+-- 计时函数
+time :: NFData a => String -> IO a -> IO a
+time label action = do
+    start <- getCurrentTime
+    result <- action
+    result' <- evaluate (force result)
+    end <- getCurrentTime
+    printf "%s took time: %.6f seconds\n" label (realToFrac (diffUTCTime end start) :: Double)
+    return result'
+
+-- 反转十进制数字（辅助函数）
+reverseDecimal :: Integer -> Integer
+reverseDecimal = go 0
+  where
+    go acc 0 = acc
+    go acc x = let (q, r) = quotRem x 10
+               in go (acc * 10 + r) q
+
+-- 十进制回文检查
+isDecimalPalindromic :: Integer -> Bool
+isDecimalPalindromic n = n == reverseDecimal n
+
+-- 二进制回文检查（使用位运算）
+isBinaryPalindromic :: Integer -> Bool
+isBinaryPalindromic n
+    | n < 0     = False
+    | otherwise = n == reverseBinary n
+  where
+    -- 计算二进制位数
+    bitLength :: Integer -> Int
+    bitLength 0 = 1
+    bitLength x = go 0 x
+      where
+        go !len 0 = len
+        go !len y = go (len + 1) (y `shiftR` 1)
+
+    -- 反转二进制位
+    reverseBinary :: Integer -> Integer
+    reverseBinary x = go 0 (bitLength x) x
+      where
+        go !acc 0 _ = acc
+        go !acc !len y
+            | testBit y 0 = go (acc `setBit` (len - 1)) (len - 1) (y `shiftR` 1)
+            | otherwise   = go acc (len - 1) (y `shiftR` 1)
+
+-- 简单方法：遍历所有数字
+mainSimple :: Integer -> IO ()
+mainSimple limit = do
+    let nums = [0..limit]
+        palindromes = filter (\n -> isDecimalPalindromic n && isBinaryPalindromic n) nums
+        result = sum palindromes
+    print result
+
+-- 生成指定长度的十进制回文数（优化版）
+palindromesOfLength :: Int -> [Integer]
+palindromesOfLength 1 = [0..9]
+palindromesOfLength len
+    | even len  = buildPalindromes evenPalindrome halfRange
+    | otherwise = buildPalindromes oddPalindrome halfRange
+  where
+    halfLen = (len + 1) `div` 2
+    start = 10^(halfLen-1)
+    end = 10^halfLen - 1
+    halfRange = [start..end]
+
+    buildPalindromes :: (Integer -> Integer) -> [Integer] -> [Integer]
+    buildPalindromes f = map f
+
+    evenPalindrome :: Integer -> Integer
+    evenPalindrome half =
+        let s = half
+            reversed = reverseDecimal s
+            shift = 10^halfLen
+        in s * shift + reversed
+
+    oddPalindrome :: Integer -> Integer
+    oddPalindrome half =
+        let s = half
+            power = halfLen - 1
+            shift = 10^power
+            reversed = reverseDecimal (s `quot` 10)
+        in s * shift + reversed
+
+-- 快速方法：生成回文数而非遍历
+mainQuick :: Integer -> IO ()
+mainQuick limit = do
+    let maxLength = length (show (limit - 1))
+        -- 限制每个长度生成的回文数不超过limit
+        limitedPalindromes = takeWhile (<= limit) $ concatMap palindromesOfLength [1..maxLength]
+        validPalindromes = filter isBinaryPalindromic limitedPalindromes
+        result = sum validPalindromes
+    print result
+
+-- 主函数
+main :: IO ()
+main = do
+    let limit = 1000000
+    putStrLn "=== mainSimple (朴素遍历法) ==="
+    time "mainSimple" $ mainSimple limit
+
+    putStrLn "\n=== mainQuick (回文数生成法) ==="
+    time "mainQuick" $ mainQuick limit

solutions/0036.Palindromes/readme.md

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+# README
+
+说起来，还是先找到所有回文数字更快，单纯计算查找确实稍慢。
+另外，就是python3.13比haskell还要快一线……我也是囧了……
+
+
+
+**haskell**
+
+```bash
+> ghc euler_36.hs
+> ./euler_36.exe
+```