Suppose we have a straightforward algorithm for a problem that does (nlog2n) steps for inputs of size n. Suppose we devise a Divide-and-Conquer algorithm that divides an input into two inputs half as big, and does D(n) = @(1) steps to divide the problem and C (n) = (na) steps to combine the solutions to get a solution to the original input. Is the Divide-and-Conquer algorithm more or less efficient than the straightforward algorithm? Justify your