CF1699E Three Days Grace

Ibti was thinking about a good title for this problem that would fit the round theme (numerus ternarium). He immediately thought about the third derivative, but that was pretty lame so he decided to include the best band in the world — [Three Days Grace](https://www.youtube.com/channel/UCJDON3eLf8709E9YfjAgLOw). You are given a multiset $ A $ with initial size $ n $ , whose elements are integers between $ 1 $ and $ m $ . In one operation, do the following: - select a value $ x $ from the multiset $ A $ , then - select two integers $ p $ and $ q $ such that $ p, q > 1 $ and $ p \cdot q = x $ . Insert $ p $ and $ q $ to $ A $ , delete $ x $ from $ A $ . Note that the size of the multiset $ A $ increases by $ 1 $ after each operation. We define the balance of the multiset $ A $ as $ \max(a_i) - \min(a_i) $ . Find the minimum possible balance after performing any number (possible zero) of operations.

N/A

N/A

In the first test case, we can apply the operation on each of the $ 4 $ s with $ (p,q) = (2,2) $ and make the multiset $ \{2,2,2,2,2,2,2\} $ with balance $ \max(\{2,2,2,2,2,2,2\}) - \min(\{2,2,2,2,2,2,2\}) = 0 $ . It is obvious we cannot make this balance less than $ 0 $ . In the second test case, we can apply an operation on $ 12 $ with $ (p,q) = (3,4) $ . After this our multiset will be $ \{3,4,2,3\} $ . We can make one more operation on $ 4 $ with $ (p,q) = (2,2) $ , making the multiset $ \{3,2,2,2,3\} $ with balance equal to $ 1 $ . In the third test case, we can apply an operation on $ 35 $ with $ (p,q) = (5,7) $ . The final multiset is $ \{6,5,7\} $ and has a balance equal to $ 7-5 = 2 $ . In the forth test case, we cannot apply any operation, so the balance is $ 5 - 1 = 4 $ .