If quadratic probing ($$h_i(k) = (H(k) + i^2) \% 11$$. Note: it

If quadratic probing ($$h_i(k) = (H(k) + i^2) \% 11$$. Note: it's not $$\pm i^2$$) is used to resolve collisions, to insert several elements, all with hash value being 2, into a hash table of size 11, then the 4th element must be placed at the position 0. ~@[](3)

答案：TRUE