the time is a little bit bad, cause cf Div.2 round 613 is right after it!

Not able to access the site. It says 502 Bad gateway.

site not working in india???

502 Bad Gateway

EXPLOSION!

503 Service Temporarily Unavailable

503 Service Temporarily Unavailable

Atcoder Not Working....... 503 Service Temporarily Unavailable

RIP AtCoder. T_T

Problem Resolved... Contest Starts Again

Is contest still rated?

2 continuous ABC(149,150) became unrated !!! lets hope rated for next rated round.

What is the problem in inputs of question D?

Quick commentary:

• A: Implementation.
• B: Implementation.
• C: $$$N \leq 8$$$ means we can generate a list of all permutations and sort them to learn their ranks, then look up $$$P$$$ and $$$Q$$$. Alternatively, we can take a permutation and count its successors (C++ builtin next_permutation) to figure out its rank.
• D: We seek $$$X \equiv \frac{1}{2} \textrm{ (mod } a_i)$$$ for all $$$i$$$, which uniquely determines $$$X$$$ modulo $$$\textrm{LCM}(a)$$$. If such an $$$X$$$ exists, it must be congruent to $$$\frac{1}{2}$$$ modulo $$$\textrm{LCM}(a)$$$. The LCM is even, so we simply want $$$\frac{\textrm{LCM}(a)}{2}$$$. We can compute and test it, being careful to avoid overflow by exiting early if the LCM exceeds $$$2M$$$ in any intermediate computation.

• E: Assume for simplicity that costs are distinct. To minimize $$$f$$$ we should modify elements in increasing order of cost. Among positions which differ between $$$S$$$ and $$$T$$$, the number of times a cost is paid is equal to the number of costs greater than it plus 1. The total number of times the $$$k^{\textrm{th}}$$$ highest cost (where $$$k$$$ is 0-indexed) is paid across all $$$(S, T)$$$ is equal to $$$2 \cdot 4^{N - k - 1} \cdot 2^k \sum_{x \leq k} (x+1)\binom{k}{x}$$$. It can be computed in constant time by using the combinatorial identity $$$\sum_{x} x\binom{k}{x} = k \cdot 2^{k-1}$$$.

• F: The value of $$$k$$$ uniquely determines $$$x = a[k] \oplus b[0]$$$. We can determine which $$$k$$$ work for each bit independently and take the intersection to get the final answer. In the 1-bit version, we want to know which rotations of $$$a$$$ are equal to either $$$b$$$ or $$$!b$$$ (corresponding to $$$x = 0$$$ and $$$x = 1$$$). We can do it by doubling $$$a$$$ and applying a string-searching algorithm (e.g. KMP) to find all occurrences of $$$b$$$ and $$$!b$$$.

Randomised Solution for F- https://atcoder.jp/contests/abc150/submissions/9403740

How to Solve D ?

In problem E sample test case 2 how is the output 124. According to my understanding, there will 8 pairs with exactly 1 difference and 4 pairs with 2 differences

So the output should be 8 * 5 + 4 * 18 = 112.

Will there be editorial in English?

I was reading the English editorial of problem D, and I noticed some mistakes there:

1) If there exist i, j such that the number of times a_i is divided by 2 and the number of times a_j is divided by 2, then ...

should be

If there exist i, j such that the number of times a_i is divided by 2 and the number of times a_j is divided by 2 is not equal, then ...

2) Let T be the least common multiple of T.

should be

Let T be the least common multiple of (a_1/2, a_2/2, ..., a_n/2).