摘要:Fancy, classic, STEP-BY-STEP Analysis strategy problem. Here is the approach: 1. Sort input arrays.. so sum[0] is a[0] * k, so we got a[0]2. Then who
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12 2017 档案
摘要:Classic problem to learn Game Theory - an advanced one: how to identify sub-games.. For every move (hit 1\2 bins) on one continuous section, the origi
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摘要:DFS + memo.
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摘要:Actually, it is a Greedy problem : )
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摘要:It is marked as Recursion on HR, but the optimal solution is DP. Step by step, discover internal mechanism.
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摘要:The trick of this problem is.. Adding 1+ coins to one pile, means nothing - you add one, then i just remove it.. so it is back to original game. So, b
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摘要:A simple NIM game in disguise: move 1 coin to one of previous slot, equals to removing.. Then no surprise, Sprague-Grundy theorem solves the problem.
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摘要:Another easy one solved by Sprague-Grundy theorem. Each pile is a sub-game, so you need to XOR SG value of all substates - since you can move # of 1 -
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摘要:A matter of OBSERVATION.. draw a triangle of parity, and watch.
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摘要:Discrete thinking - Play & Observe! " you can move numbers in any quadrant at the following positions to [x][y]: [x][y] or // upper left quadrant [x][
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摘要:Another intuitive of application of Sprague-Grundy theorem.
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摘要:An intuitive problem for learning & applying Sprague-Grudy theorem: https://zhuanlan.zhihu.com/maigo/20611132 Lesson learnt, in Sprague-Grundy theorem
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摘要:The major trick is from another similar HR problem: subarray with max XOR: Build a binary tree bit by bit, and go from MSB to LSB, greedily. In this o
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摘要:Fun Greedy. My first thought was a DFS based solution... however the editorial provides a super neat 2-pass O(n) solution: Pass 1: if s[l] != s[r], ch
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摘要:Over-thinking will not get you there though : ) The trick is QUITE easy: # of unique chars.
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