By Hadras, history, 7 weeks ago,

Can this problem be solved using Binary Search

• +4

 » 7 weeks ago, # |   0 UP
 » 7 weeks ago, # |   0 NO. First test case, 100 can be represented with 1 banknote. And so does 1. Basically, you cannot compare it to the final answer.
•  » » 7 weeks ago, # ^ |   0 What do u mean by final answer
•  » » » 7 weeks ago, # ^ |   0 For each test case, print one integer — the minimum positive number of burles s that cannot be represented with k or fewer banknotes.
•  » » 7 weeks ago, # ^ |   0 Thanks
 » 7 weeks ago, # | ← Rev. 4 →   0 You can use binary search in the array a so that you can say that the cost(10^b) = 10^(b-last) where last is the last ai such that ai <= b and use this to solve the problem(this is quite useless since n = 10 but actually optimize it 157598538), otherways I don't where to use binary search. If you meant binary search the answer, it is not possible since the cost(x) does not strictly increment, in a case like a: {0, 4} cost(10000) = 1, cost(9999) = 9999.
•  » » 7 weeks ago, # ^ |   0 Thanks
 » 7 weeks ago, # |   0 No. Binary search requires the searchspace to have a special property. If it can be done for some value x then it can be done also for x+1 and it's enough.(The direction can be reversed, ie x -> x-1).This problem clearly doesn't satisfy this. 9: 9notes 10: 1 note. (decreased) 11: 2 notes (increased) 
•  » » 7 weeks ago, # ^ |   0 Thanks