Question

I need to create a c++ print function that takes a vector of values ex: <1 2 3 4 5 6> and prints it in the correct way with the given dimensions. Example 2x3 = [[1,2,3],[4,5,6]] or 3x2 = [[1,2],[3,4],[5,6]]. To make this even more difficult there can be 3 dimensional matrixs. For example for <1 2 3 4 5 6 7 8 9 10 11 12> with dimensions depthXheightXwidth: 2x3x2 = [[[1,2],[3,4],[5,6]],[[7,8],[9,10],[11,12]]]. The depth height and width could be any value as well as the numbers in the arrays so I need a function that can handle all cases. Any help would be greatly appreciated.

Answer 1

Well, this is actually quite a difficult problem as you want to handle all possible cases for any count of elements. And that can take very higher complexity as it may raise a case of combinatorial explosion. But to find at least a way to represent the given set of elements in an array form either in 2D or 3D from, there is a solution. In the following, I'm writing an algorithm to perform the given task. I'm leaving the task of writing the whole code to you, and giving the key idea to use in that code Algo //this could be the main driver program of C++ which takes input from the stdin, i.e. keyboard 1. Take a line as input <getline (cin, line) is available, line is of type string 2. Ignore the first character) by using redirectionover line to some integer 3. While (this integer value is not> i.e. the ending char) a. Count the elements entered as count 4. Now, start dividing(save remainder as well) count by 2 onward after knowing that it is not prime number a. If there is no remainder then it seems you have got a candidate < divisor, quotient b. Further divide that quotient by 2 onward if that is not prime number. If you get any case with no remainder means, you have got another solution. <divisorl, quotient lor divisor2, quotient2> Do step 4 until the final quotient is surely a prime number Done Now to check if a number is prime no, you can simply make another module/function which will divide the given no with 2 up to the square root of that no to check if there is any remainder at any case. If found any remainder means it is not prime, hence you can divide it to find another dimension which gives another solution to the desired problem