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2014 Asian ACM/ICPC 牡丹江赛区题目. 题目的大意是,Bob同学和其他N – 1个人考试被分在A组,同时有M个人被分到了B组,给定除了bob以外所有的人的成绩。 由于公式题目已经给了,所以唯一难点就是当p趋近于0 的时候,lim(p*logb(p) 的数值了…其实就是0。 逐个状态隐式的包含了,你下了多少个棋子i * j – k。 Trick 1. 如果表达式中有operator,一定要保证最后一个字符是operator,如果不是的话第一步是把operator交换到最后。 要点2 dfs json tree hash 每个node的key, query也用hash值去match cache的key. 在树上找两个fire station x, y,使得任何个点 到fire station的距离的最大值最小. 因为树一定会被这个连个点切成两颗子树,我们必定选择树的直径l路径的终点,否则答案必定大于等于(l 1) / 2. 同理每棵子树上的点,也必定选择子树直径的中点,否则答案大于等于(lsubstr 1) / 2 不小于最终答案。 这样才能保证,各层的索引个数为N * (1/p) h。

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ronaflx | ronaflx.com Reviews
<META>
DESCRIPTION
2014 Asian ACM/ICPC 牡丹江赛区题目. 题目的大意是,Bob同学和其他N – 1个人考试被分在A组,同时有M个人被分到了B组,给定除了bob以外所有的人的成绩。 由于公式题目已经给了,所以唯一难点就是当p趋近于0 的时候,lim(p*logb(p) 的数值了…其实就是0。 逐个状态隐式的包含了,你下了多少个棋子i * j – k。 Trick 1. 如果表达式中有operator,一定要保证最后一个字符是operator,如果不是的话第一步是把operator交换到最后。 要点2 dfs json tree hash 每个node的key, query也用hash值去match cache的key. 在树上找两个fire station x, y,使得任何个点 到fire station的距离的最大值最小. 因为树一定会被这个连个点切成两颗子树,我们必定选择树的直径l路径的终点,否则答案必定大于等于(l 1) / 2. 同理每棵子树上的点,也必定选择子树直径的中点,否则答案大于等于(lsubstr 1) / 2 不小于最终答案。 这样才能保证,各层的索引个数为N * (1/p) h。
<META>
KEYWORDS
1 ronaflx
2 acmer cser algorithm爱好者,学术水平有待提高
3 1 response
4 出差期间,闲来无事,把这套题过了一边
5 我力所能及的题目有6道吧,都及时a掉了
6 其中四道题都是1a
7 虽然速度已经是大不如前,但是试错狂魔的我,能有4个1a也很欣慰了
8 总体来讲这六道题的难度分布还不错,代码量也很足
9 zoj出品还是有保障的
10 average score
CONTENT
Page content here
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PAGE
ronaflx,acmer cser algorithm爱好者,学术水平有待提高,1 response,出差期间,闲来无事,把这套题过了一边,我力所能及的题目有6道吧,都及时a掉了,其中四道题都是1a,虽然速度已经是大不如前,但是试错狂魔的我,能有4个1a也很欣慰了,总体来讲这六道题的难度分布还不错,代码量也很足,zoj出品还是有保障的,average score,是一道枚举题,如果bob去了b组,两组人的平均分都会提高,求bob的分数的最小和最大的可能值,information entropy,开学近况
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ronaflx | ronaflx.com Reviews

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2014 Asian ACM/ICPC 牡丹江赛区题目. 题目的大意是,Bob同学和其他N – 1个人考试被分在A组,同时有M个人被分到了B组,给定除了bob以外所有的人的成绩。 由于公式题目已经给了,所以唯一难点就是当p趋近于0 的时候,lim(p*logb(p) 的数值了…其实就是0。 逐个状态隐式的包含了,你下了多少个棋子i * j – k。 Trick 1. 如果表达式中有operator,一定要保证最后一个字符是operator,如果不是的话第一步是把operator交换到最后。 要点2 dfs json tree hash 每个node的key, query也用hash值去match cache的key. 在树上找两个fire station x, y,使得任何个点 到fire station的距离的最大值最小. 因为树一定会被这个连个点切成两颗子树,我们必定选择树的直径l路径的终点,否则答案必定大于等于(l 1) / 2. 同理每棵子树上的点,也必定选择子树直径的中点,否则答案大于等于(lsubstr 1) / 2 不小于最终答案。 这样才能保证,各层的索引个数为N * (1/p) h。

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2014 Asian ACM/ICPC 牡丹江赛区题目. 题目的大意是,Bob同学和其他N – 1个人考试被分在A组,同时有M个人被分到了B组,给定除了bob以外所有的人的成绩。 由于公式题目已经给了,所以唯一难点就是当p趋近于0 的时候,lim(p*logb(p) 的数值了…其实就是0。 逐个状态隐式的包含了,你下了多少个棋子i * j – k。 Trick 1. 如果表达式中有operator,一定要保证最后一个字符是operator,如果不是的话第一步是把operator交换到最后。 要点2 dfs json tree hash 每个node的key, query也用hash值去match cache的key. 在树上找两个fire station x, y,使得任何个点 到fire station的距离的最大值最小. 因为树一定会被这个连个点切成两颗子树,我们必定选择树的直径l路径的终点,否则答案必定大于等于(l 1) / 2. 同理每棵子树上的点,也必定选择子树直径的中点,否则答案大于等于(lsubstr 1) / 2 不小于最终答案。 这样才能保证,各层的索引个数为N * (1/p) h。

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