### hulk_baba's blog

By hulk_baba, history, 5 years ago,

Please look at problem here. I figured out how to do it: 1. I will check among all possible quasibinary numbers which provides me least numbers of such numbers whose sum is equal to n(the given number) 2. I memoize the solution. Basic DP stuff. 3. In all my tests I got correct answer for number of numbers but I can't construct the solution i.e actual numbers which form the solution. Can you help me with this.

I have submitted my solution here

• +4

 » 5 years ago, # |   0 Auto comment: topic has been updated by hulk_baba (previous revision, new revision, compare).
 » 5 years ago, # |   0 Still No answers?
 » 5 years ago, # | ← Rev. 3 →   0 Save curr in an array and then trace back the numbers that you chose. Something likeint cur = n;while(cur){ans.push_back(saved[cur]);cur -= saved[cur];}
•  » » 5 years ago, # ^ |   0 Thanks. That was very helpful. got it right.
 » 5 years ago, # | ← Rev. 3 →   0 Do smth like thatint powten = 1; `for(int i=0;i<7;i++){ int x = (n / powten) % 10; for(j=0;j
•  » » 5 years ago, # ^ |   0 how does that work? What is the algorithm?