Difference between revisions of "2018 Multi-University, Nowcoder Day 4"
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Upsolved by kblack. | Upsolved by kblack. | ||
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+ | 题意:一堆概率出现的点,求在任意点左下角的点数的期望。 | ||
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+ | 题解:记录不在任意点左下角的概率,每个点相当于对一个矩形乘以 $(1-p)$,扫描线+线段树即可。 | ||
== Problem F == | == Problem F == |
Revision as of 11:06, 28 July 2018
Problem A
Solved by zerol. 02:17 (+2)
Problem B
Problem C
Solved by ultmaster. 02:57 (+1)
Problem D
Solved by zerol. 01:19 (+1)
Problem E
Upsolved by kblack.
题意:一堆概率出现的点,求在任意点左下角的点数的期望。
题解:记录不在任意点左下角的概率,每个点相当于对一个矩形乘以 $(1-p)$,扫描线+线段树即可。
Problem F
Solved by ultmaster. 00:37 (+)
Problem G
Solved by kblack. 00:39 (+)
枚举即可,温暖的规律签到。
Problem H
Problem I
Problem J
Solved by kblack. 04:05 (+8)
题意:naive 的 hash 函数,naive 的解决方法(往后放),给出结果,求字典序最小的原插入顺序。
题解:很明显的拓扑排序,原题解是线段树优化。也可以考察特殊条件,所有相交的区间合并,包含的区间只会被跳一次,瞎几把搞一搞就好了,注意无解的判断即可。