제출 #622281

#	제출 시각UTC-0	아이디	문제	언어	결과	실행 시간	메모리
622281	2022-08-04 05:48:41	jophyyjh	Lottery (CEOI18_lot)	C++14	65 / 100	78 ms	21376 KiB

lot

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/**
 * Notes during contest.
 * 
 * ------ A ------
 * Looks like a dp.
 * 
 * ------ B ------
 * I think i've seen sth similar on luogu. First, let's assume that d >= 0 and i'll
 * use the words "increase" & "decrease". If we wanna increase an interval by d, we
 * can greedily increase a suffix (instead of just an interval in the middle). If we
 * are to decrease an interval by d, we can greedily decrease a prefix. The two cases
 * are symmetric, so we can assume that one always increase a suffix by 0 <= d <= x.
 * And, if we're increasing a suffix, why don't we just do d=x? The rest is quite
 * straight-forward.
 * 
 * ------ C ------
 * For k_j = 0, we have to find the num of times each interval appeared. This can be
 * effectively done with str hashing. [S3] solved. [S1] is just brute-force: we can
 * do a O(n^2) for loop, iterating over all pairs of starting pos, naively comparing
 * the dist. of 2 substr. [S2] is a O(n^2) comparison between pairs of VALUES and
 * apply a difference array.
 * We're only looking for the num of mismatches. Let's compress the values (a_i:
 * 10^9 -> 10^4).
 * 
 * Time Complexity 1: O()
 * Time Complexity 2: O(n * log(n))
 * Time Complexity 3: O(n^2 + q) ([S1-2]), O(n)    (non-deterministic hashing)
 * Implementation 1         (Just for partials, [S1-2], [S3]. [S3] not finished)
*/
 
 
 
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• Subtask 1: 25 / 25
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 0 / 15
• Subtask 5: 0 / 20