fix: fix comment for new added files.

- translate zh-CN comment into en-US.
- change some comment into Doxygen style.
- add lost Doxygen comment.
- enrich the testbench for ceil_div.
- add lost metaprogramming functions for some files in macro namespace.

src/yycc/num/op.hpp

@@ -3,32 +3,35 @@
 #include <concepts>
 
 /**
- * @brief The namespace providing function relative robust numeric operations.
+ * @brief The namespace providing functions for robust numeric operations.
  * @details
- * 在使用Rust编写一些程序后，深刻地认识到Rust中针对基本类型的运算符的丰富程度。
- * 提供了诸如整数向上取整除法等便利操作。
- * 因此我在这个命名空间里复刻Rust中的这些便利功能。
+ * After writing some programs in Rust, I deeply appreciated the richness of operators
+ * for primitive types in Rust, which provides convenient operations like ceiling integral division.
+ * Therefore, I replicate these convenient features from Rust in this namespace.
  *
- * 由于没有需求，目前暂未实现以下功能：
- * \li 仅支持无符号整数向上取整除法。
- * \li 显式指定运算溢出行为
+ * Currently unimplemented features due to lack of demand:
+ * \li Only supports unsigned integer ceiling division
  */
 namespace yycc::num::op {
 
     /**
-     * @brief 无符号整数的向上整除
+     * @brief Unsigned integer ceiling division
      * @details
-     * 执行两个无符号整数之间的整除，并将结果向上取整。
-     * 该函数可能在很多加密函数中进行使用以分配合适空间。
-     * @tparam T 除法操作建立在的无符号整数类型
-     * @param[in] lhs 左操作数
-     * @param[in] rhs 右操作数
-     * @return 向上取整的整除结果
+     * Performs division between two unsigned integers and rounds up the result.
+     * @exception std::logic_error If the divisor is zero
+     * @tparam T The unsigned integer type for division operation
+     * @param[in] lhs Left operand
+     * @param[in] rhs Right operand
+     * @return Ceiling division result
      */
     template<typename T>
         requires std::unsigned_integral<T>
     T div_ceil(T lhs, T rhs) {
+        // Check divisor first
         if (rhs == 0) throw std::logic_error("div with 0");
+        // YYC MARK:
+        // We use this algorithm, instead of traditional `(lhs + rhs - 1) / rhs`, 
+        // which may have unsafe overflow case.
         return (lhs % rhs == 0) ? (lhs / rhs) : (lhs / rhs) + 1u;
     }