Codeforces Round 479 (Div. 3)

5 problems from Codeforces Round 479 (Div. 3) (contest 977), difficulty 800-1700. 3/5 solutions verified against sample I/O.

5 items

Codeforces Round 479 (Div. 3)

Div. 3 | 5 problems | 3/5 verified | Difficulty 800-1700 | 6m 59s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Wrong Subtraction	800	implementation	217,381	1m 11s	✓
B	Two-gram	900	implementation, strings	49,745	1m 15s
C	Less or Equal	1200	sortings	43,322	1m 33s	✓
D	Divide by three, multiply by two	1400	dfs-and-similar, math, sortings	20,523	1m 26s	✓
F	Consecutive Subsequence	1700	dp	25,119	1m 34s

Practice › Codeforces Solutions › Codeforces Round 479 (Div. 3) ›

CF 977F - Consecutive Subsequence

We are given a sequence of integers in the order they appear. From this sequence we want to pick some elements while preserving order, but we are only allowed to keep a subsequence that looks like a run of consecutive integers increasing by exactly one each step, such as $x…

Practice › Codeforces Solutions › Codeforces Round 479 (Div. 3) ›

CF 977D - Divide by three, multiply by two

We are given a multiset of integers that originally came from a single starting value that Polycarp repeatedly transformed. Each transformation either doubles the current number or divides it by three when it is divisible.

Practice › Codeforces Solutions › Codeforces Round 479 (Div. 3) ›

CF 977C - Less or Equal

We are given a list of integers and asked to construct a value $x$ between 1 and $10^9$ such that exactly $k$ elements of the list are less than or equal to $x$. If no such value exists, we must output $-1$.

Practice › Codeforces Solutions › Codeforces Round 479 (Div. 3) ›

CF 977B - Two-gram

We are given a string of uppercase English letters, and we are asked to look at every adjacent pair of characters in it. Each such adjacent pair forms a “two-letter pattern”, for example the string “ABAC” contains “AB”, “BA”, and “AC”.

Practice › Codeforces Solutions › Codeforces Round 479 (Div. 3) ›

CF 977A - Wrong Subtraction

The task describes a very specific way of reducing a positive integer repeatedly. Instead of simply subtracting one in the usual arithmetic sense, the operation depends on the last digit of the number.