X = 712360810625491574981234007851998使用链表表示，每个节点都是一个unsigned int

有没有快速的方法X << 8罢工>X << 591除了之外X * 2^8罢工>X * 2^591 ?

任意位数的位移非常容易。只要记住将溢出的位移至下一个元素。就这样

下面是左移 3 的示例

uint64_t i1, i2, i3, o1, o2, o3; // {o3, o2, o1} = {i3, i2, i1} << 3;

o3 = i3 << 3 | i2 >> (32 - 3);
o2 = i2 << 3 | i1 >> (32 - 3);
o1 = i1 << 3;

与右移类似，只是反向迭代。

Edit:

It seems that you're using base 10⁹ for your large number, so binary shifting does not apply here. "Shifting" left/right N digits in a base B is equivalent to multiplying the number by B^N and B^-N respectively. You can't do binary shift in decimal and vice versa

If you don't change your base then you have only one solution, that's multiplying the number by 2⁵⁹¹. If you want to shift like in binary you must change to a base that is a power of 2 https://stackoverflow.com/q/23840565/995714 like base 2³² or base 2⁶⁴

A general solution to shift would be like this, with the limbs ("digits" or each small word unit in a big integer, in arbitrary-precision arithmetic term) stored in little-endian and each digit is in base 2^{CHAR_BIT*sizeof(T)}

template<typename T,
    class = typename std::enable_if<std::is_unsigned<T>::value>::type>
void rshift(std::vector<T>& x, std::size_t shf_amount) // x >>= shf_amount
{
    // The number of bits in each limb/digit
    constexpr std::size_t width = CHAR_BIT*sizeof(T);
    if (shf_amount > width)
        throw; // or zero out the whole vector for saturating shift

    // Number of limbs to shift
    const std::size_t limbshift = shf_amount / width;
    // Number of bits to shift in each limb
    const std::size_t shift     = shf_amount % width;
    
    std::size_t i = 0;
    // Shift the least significant bits
    for (; i < x.size() - limbshift - 1; ++i)
        x[i] = (x[i +     limbshift] >> shift) |
               (x[i + 1 + limbshift] << (width - shift));
    x[i] = x[i + limbshift] >> shift;
    i++;
    // Zero out the most significant bits
    for (; i < x.size() ; ++i)
        x[i] = 0;
}

此外，从标签来看，您可能使用链接列表来存储肢体，由于元素分散在内存空间中，该链接列表对缓存不友好，并且由于next指针。实际上，在这种情况下，指针和内存分配使用的内存甚至比用于存储数据位的内存还要大。事实上，在大多数现实生活问题中你不应该使用链表

• Bjarne Stroustrup 说我们必须避免链表 https://stackoverflow.com/q/34170566/995714
• 为什么你永远、永远、永远不要再在你的代码中使用链表 https://kjellkod.wordpress.com/2012/02/25/why-you-should-never-ever-ever-use-linked-list-in-your-code-again/
• 数字运算：为什么你不应该再在代码中使用链表 https://www.codeproject.com/articles/340797/number-crunching-why-you-should-never-ever-ever-us
• Bjarne Stroustrup：为什么应该避免使用链表 https://www.reddit.com/r/programming/comments/25xpre/bjarne_stroustrup_why_you_should_avoid_linked/
• 列表是邪恶的吗？——Bjarne Stroustrup https://isocpp.org/blog/2014/06/stroustrup-lists