Difference between revisions of "2018-2019 ACM-ICPC, Asia Dhaka Regional Contest"
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不过还有一个结论感觉很对但是不会证明,就是经过一个轮换之后积木形状相同。 | 不过还有一个结论感觉很对但是不会证明,就是经过一个轮换之后积木形状相同。 | ||
+ | |||
+ | 时间复杂度 $O(n^2)$。 | ||
== Problem I == | == Problem I == |
Revision as of 11:50, 19 October 2020
Problem A
Unsolved.
Problem B
Solved by yanghong. 04:34 (+)
Problem C
Solved by bingoier. 00:35 (+)
Problem D
Unsolved.
Problem E
Solved by bingoier. 01:17 (+1)
Problem F
Solved by yanghong. 01:06 (+)
Problem G
Unsolved.
Problem H
Solved by Once. 03:08 (+2)
大概可以猜测一下,一个积木盘只有绕一圈之后才会出现一个轮换。所以我们就暴力把原来的积木标号之后转一圈,得到所有轮换,求出轮换的最小公倍数就是答案。然而模数不是素数,所以答案需要处理出所有的质因子后暴力相乘。
不过还有一个结论感觉很对但是不会证明,就是经过一个轮换之后积木形状相同。
时间复杂度 $O(n^2)$。
Problem I
Unsolved.
Problem J
Solved by bingoier. 00:07 (+)