<algorithm> function for finding last item less-than-or-equal to, like lower_bound

In a sorted container, the last element that is less than or equivalent to x, is the element before the first element that is greater than x.

Thus you can call std::upper_bound, and decrement the returned iterator once. (Before decrementing, you must of course check that it is not the begin iterator; if it is, then there are no elements that are less than or equivalent to x.)

Here is a wrapper function around upper_bound which returns the largest number in a container or array which is less than or equal to a given value:

template <class ForwardIterator, class T>
  ForwardIterator largest_less_than_or_equal_to ( ForwardIterator first, 
                                                  ForwardIterator last,
                                                  const T& value)
{
  ForwardIterator upperb = upper_bound(first, last, value);

  // First element is >, so none are <=
  if(upperb == first)
    return NULL;

  // All elements are <=, so return the largest.
  if(upperb == last)
    return --upperb;

  return upperb - 1;
}

For a better explanation of what this is doing and how to use this function, check out:

C++ STL — Find last number less than or equal to a given element in an array or container

I have test your reverse iterator solution, it is correct.

Given v is sorted by '<'

Find last element less than x:

auto iter = std::upper_bound(v.rbegin(), v.rend(), x, std::greater<int>());
if(iter == v.rend())
    std::cout<<"no found";
else
    std::cout<<*iter;

Find last element less than equal x:

auto iter = std::lower_bound(v.rbegin(), v.rend(), x, std::greater<int>());
if(iter == v.rend())
    std::cout<<"no found";
else
    std::cout<<*iter;

This is better than iter -= 1 version