Guys, please answer this, it won't take much time for those who have already solved this.
http://mirror.codeforces.com/contest/552/problem/C
In this problem, if X ≤ wk for where k is the largest possible, then we don't need to use all coins that have higher power than k + 1, i.e. coins wk + 2, wk + 3, ...wn will not be used.
To start with the proof, I will take wk + 2 first.
I need to prove that wk + 2 is not used in all of the valid solutions which means, wk + 2 doesn't occur either on the left or right side of the equation shown at the top. Now, if I could somehow prove that using wk + 2 in left side or right side, I cannot arrive at a solution I would be complete with my proof. I will first put wk + 2 in the left (along with X) and see. I have X + wk + 2 on the left, I also know that X ≤ wk. I can't work any further.
I am not able to prove how.
Ok, guys I got it!!!! (someone helped me). Here's a more easy proof.
Lets think a different way. What are the coin changes that I can make using wk + 2. What is the smallest such change? The smallest change is equal to when we put wk + 2 on the left and all smaller weights on the right i.e.
wk + 2 - wk + 1 - wk - .... - w0
Unable to parse markup [type=CF_TEX]
— 1)/(w - 1)But this amount is greater than wk.
Thus, if we use wk + 2, the change must be greater than wk. But, X ≤ wk. So, we can't make any such X. Proof is complete.