Codeforces Round #587 (Div. 3) Unofficial Editorial

Правка en3, от LiM_256, 2019-09-22 06:21:52
• A

It is easy to figure out that any $s[i]$ and $s[i - 1]$ which $i$ is even ($1$ index) must have same ammount of 'a' and 'b's. So just change it greedily.

Solution
• B

We don't need to consider "$+1$", because it is a constant. Without it, we can sort the array undecreasing and just implement it one by one. This greedy algorithm can be proved with rearrangement inequality.

Solution
• C

You can solve this problem intuitively, with checking each "uncovered" point by trying $8$ directions of $(x1, y1)$ with 0.5 length, which x1 is in x[1...6], y1 is in y[1...6].

Solution
• D

It is easy to figure that $z$ must be divided by $abs(a[i] - a[i - 1])$. So $z$ will become maximum if we take $z$ as its greatest common divisior, and y will become minimum.

Solution
• E

We can do two binary search and solve the problem easily, and check answer with simple math formula.

Solution
• F

We can construct such DP method:

DP

We can maintain it using simple data structure, such as segment tree. And the problem can be solved.

Solution

#### История

Правки

Rev. Язык Кто Когда Δ Комментарий
en3 LiM_256 2019-09-22 06:21:52 59
en2 LiM_256 2019-09-22 06:18:48 5 Tiny change: 'e it greedy.\n\n<spo' -> 'e it greedily.\n\n<spo'
en1 LiM_256 2019-09-22 06:17:56 6654 Initial revision (published)