Codeforces Round #281 (Div. 2)_html/css_WEB-ITnose

This question is not difficult either.

But I keep making mistakes. Either I misread the question or the array is too small.

A, B, C, and D are all pretty easy. E is actually quite simple, but I didn’t understand it at the time. .

If C, just enumerate all possible d, and the complexity is sorted nlogn

D , For odd numbers, Black only needs to move symmetrically with White to win

For even numbers, White takes one step to 1, 2 and it becomes an odd number situation, and then Black moves, White just needs to move symmetrically. So in the end White will definitely win

E

For the given t, a, b

We first judge the special Done,

It’s nothing more than the case of t = 1, a=1

Special judgment based on whether b is equal to 1

And then the rest The situation depends on the equation

a0 a1t a2t^2 ...=a

a0 a1a a2a^2 ...=b

Then move the terms to get

a1 a2t a3t^2 ...= (a-a0)/t

a1 a2a a3a^2 ...= (b-a0) /a

You will find that this problem can be solved recursively.

The value of a0 here has requirements

(a-a0) %t ==0

(b-a0)%a==0

In other words, a0%a == b % a, a0 % t == a % t

Then we found that the amount of enumerating a0 is actually very small

For a0%a == b %a has a0= k * a b �

a0 <= b && a0 <= a

will find k=0 or 1, and it must satisfy a0 % t == a % t

Then the next step is recursion. Here’s the answer

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