#include <vector> std::vector<long int> as; long int a(size_t n){ if(n==1) return 1; if(n==2) return -2; if(as.size()<n+1) as.resize(n+1); if(as[n]<=0) { as[n]=-4*a(n-1)-4*a(n-2); } return mod(as[n], 65535); } The above code sample using memoization to calculate a recursive formula based on some input n. I know that this uses memoization, because I have written a purely recursive function that uses the same formula, but this one much, much faster for much larger values of n. I’ve never used vectors before, but I’ve done some research and I understand the concept of them. I understand that memoization is supposed to store each calculated value, so that instead of performing the same calculations over again, it can simply retrieve ones that have already been calculated.
My question is: how is this memoization, and how does it work? I can’t seem to see in the code at which point it checks to see if a value for n already exists. Also, I don’t understand the purpose of the if(as[n]<=0). This formula can yield positive and negative values, so I’m not sure what this check is looking for.
Thank you, I think I’m close to understanding how this works, it’s actually a bit more simple than I was thinking it was.
I do not think the values in the sequence can ever be 0, so this should work for me, as I think n has to start at 1.
However, if zero was a viable number in my sequence, what is another way I could solve it? For example, what if five could never appear? Would I just need to fill my vector with fives?
Edit: Wow, I got a lot of other responses while checking code and typing this one. Thanks for the help everyone, I think I understand it now.
if (as[n] <= 0)is the check. If valid values can be negative like you say, then you need a different sentinel to check against. Can valid values ever be zero? If not, then just make the testif (as[n] == 0). This makes your code easier to write, because by default vectors ofints are filled with zeroes.