Difference between revisions of "2018 Multi-University, HDU Day 6"
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== Problem I == | == Problem I == | ||
+ | Solved by kblack. 01:01 (+1) | ||
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+ | 题意:村民狼人互相指认,村民一定说实话,狼人可以胡说八道,求铁狼数量。 | ||
+ | |||
+ | 题解:显然,可能所有人都在胡说八道。如果指认组成一个圈,其中一项指控是狼人其他是村民,那么狼人指控的目标一定是铁狼,认为铁狼是村民的也是铁狼,也要计算在内。 | ||
== Problem J == | == Problem J == | ||
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== Problem K == | == Problem K == |
Revision as of 11:41, 8 August 2018
ultmaster: 除了我签了个到,全是 kblack 做的。可能 kblack 1v3 结局也差不多吧。。。
Problem A
Solved by ultmaster. 00:39 (+4)
题意:小学积分题。
题解:不会积分,上了个 Wolfram Alpha。没想到漏除一个 $b$(纸上是有的,抄上去就没有了)。对着一个明显不对的答案调了好久,还仔细考虑有没有精度问题。答案是 $\pi a + 2b$。注意输出要去尾。根本不需要什么 long double...
Problem B
Solved by kblack. 02:12 (+2)
Problem C
Solved by kblack. 04:12 (+5)
Problem D
Problem E
Problem F
Problem G
Unsolved. (-6)
Problem H
Problem I
Solved by kblack. 01:01 (+1)
题意:村民狼人互相指认,村民一定说实话,狼人可以胡说八道,求铁狼数量。
题解:显然,可能所有人都在胡说八道。如果指认组成一个圈,其中一项指控是狼人其他是村民,那么狼人指控的目标一定是铁狼,认为铁狼是村民的也是铁狼,也要计算在内。
Problem J
Problem K
Upsolved by ultmaster. (-5)
题意:$M_n(i,j) = 1 \text{ if} \binom{i}{j} \bmod p > 0 \text{ else } 0$. $F(n,k) = \sum_{i = 0}^{p^n-1}\sum_{j=0}^{p^n-1}M_n^k(i,j)$. 求 $(\sum_{n=1}^N\sum_{k=1}^K F(n,k)) \bmod (10^9 + 7)$.
题解:打表找规律,最后发现如果枚举 $k$ 的话,是一个首项为 $q=p(p+1)\cdots(p+k)/(k+1)!$,公比也为 $q$ 的等比数列。求前 $N$ 项和即可。
ultmaster: 小学生常犯错误:等比数列求和直接套公式,从来不考虑公比为 $1$ 的情况,到最后都没看出来。(差点就有贡献了啊。。。心痛。。。
Problem L
Solved by kblack. 00:23 (+)
温暖的签到。