How to solve E ?

can't we do it via dp ?

My solution for E:

Among the first 2(X+Y) seconds, the train stops at Y of those seconds. Label these seconds 0,1,2...Y-1. Also, if second i is labeled, apply the same label to second i+2(X+Y).

Similarly, among the first P+Q seconds, label the seconds where Takahashi is awake, and if second i is labeled, apply the same label to second i+P+Q.

The problem is to find the earliest second that was simultaneously labelled in both steps. We can brute-force this by enumerating all Y*Q pairs of labels and considering when is the earliest second labelled with those specific pairs of labels. For each pair, the answer is an intersection of APs (https://math.stackexchange.com/questions/1378976/computing-the-intersection-of-two-arithmetic-sequences-a-mathbbz-b-cap).

For D I simulated all possible card selections (2 for loops, skipping invalid combinations that pass the card limit)

then calculating the sums after picking those cards

if its bigger add number of ways to get that card combination to num(if the cards picked by both are the same, let cnt = num of remaining cards with i written on them not picked. Its cnt * cnt-1 ways,

else its num of remaining cards numbered i * number of remaining cards numbered j) Add the same to tot then output num/tot.

btw i used doubles. You can check my solution here:

https://atcoder.jp/contests/abc193/submissions/20554741

Note that all numbers usable as $$$a$$$ are smaller or equal to sqrt(n), so max ~1e5.

Then we simply find all numbers (collect them in a set) we can build by all possible $$$a^b$$$, and subtract this number from n.

run two for loops i and j to calculate all i^j <= N, storing values in a set,

the first loop runs from 2 to sqrt(N), while the second runs starting from 2 till i^j > N (which is in worst case logN)

ans = N — Set size

https://atcoder.jp/contests/abc193/submissions/20535157

works in O(log N * sqrt(N))

As you can see clearly, the maximum base number is not more than $$$10^5$$$, and the maximum power does not exceed to $$$\left\lfloor\log_{2}10^{10}\right\rfloor=33$$$. So we just directly list all numbers within n that satisfy the b-th power form of a, and then remove the multiplicity then get the answer :)

Code using set
Code using map

Hope it'll help you :)

Yes best round in recent history. Hope this becomes a trend!