Please subscribe to the official Codeforces channel in Telegram via the link: https://t.me/codeforces_official. ×

### aMitkvikram's blog

By aMitkvikram, history, 7 months ago, ,

## 977D — Divide by three, multiply by two

1. We can see that all numbers in given sequence are distinct. since all numbers in sequence are of the form . Hence if ai = aj then = = 2x - m = 3y - n which is not possible because any power of 2 will be an even number and any power of 3 will be an odd number.

2. if there exists in sequence then 2 * ai can not be in sequence and vice versa. We can prove it using contradiction. Suppose there is a number ai such that both and 2 * ai exists in sequence. by little bit of tricks this = , this again is not possible by same argument as above, we just have to change the order of exponents.

3. Hence for each ai in sequence we see if or 2 * ai is present(remember that only one of them can be present). Now if there is any ai such that both and 2 * ai is not in sequence then this should be an. if there is any such ai that for all 0 ≤  j ≤ n:   ≠  ai AND 2 * ai  ≠  ai, then this is a1.

4. Once you have got a1, you keep on producing sequence just by doing binary search for and 2 * ai (remember only one of them is present so once you find it you print it).

•
• +3
•

 » 7 months ago, # |   0 Auto comment: topic has been updated by aMitkvikram (previous revision, new revision, compare).
 » 6 months ago, # |   +1 Can you explain your 1st statement ?
•  » » 6 months ago, # ^ |   0 I have updated the first statement. Note that any power of 2 will be even and any power of 3 will be odd, hence (power of 2) can not be equal to (power of 3).
 » 6 months ago, # |   0 Auto comment: topic has been updated by aMitkvikram (previous revision, new revision, compare).
 » 3 months ago, # |   0 https://codeforces.com/contest/977/submission/38108720 I wrote a simple comparator function to sort the array. Math is similar.
•  » » 3 months ago, # ^ |   0 There are many ways to solve this problem but your solution is concise. I solved it using dfs.