Difference between revisions of "ACM-ICPC 2018 Xuzhou Online Contest"
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Solved by zerol. 02:51 (+6) | Solved by zerol. 02:51 (+6) | ||
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+ | 题意:求模 2 意义下很多次二维卷积的结果。 | ||
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+ | 题解:类似快速幂,但需要求出一次卷积的效果(每一个位置由所有位置中若干个位置 1 的个数的奇偶性唯一决定),总复杂度 $O(n^4 \log t)$。 |
Revision as of 09:59, 9 September 2018
ECNU Foreigners
Problem A
Solved by ultmaster. 01:19 (+)
Problem B
Solved by kblack. 01:16 (+1)
Problem C
Solved by ultmaster. 02:56 (+2)
Problem D
Solved by zerol. 00:59 (+2)
题意:求 $\sum_{i=1}^m \mu(in)$
题解:如果 $n$ 中有平方因子,那么显然答案是 0。否则相当于求 $\mu(n) \cdot \sum_{i=1}^m \mu'(i)$,其中 $\mu'$ 是在 $\mu$ 的基础上不把 $n$ 的质因数当质数。类似于求 $\mu$ 的前缀和的方法(任意一种亚线性筛),改一改就能过了。
Problem E
Problem F
Solved by ultmaster. 00:26 (+)
Problem G
Solved by kblack. 00:34 (+)
Problem H
Solved by zerol. 01:16 (+)
Problem I
Solved by kblack. 00:16 (+)
Problem J
Solved by kblack. 02:42 (+3)
Problem K
Solved by zerol. 02:51 (+6)
题意:求模 2 意义下很多次二维卷积的结果。
题解:类似快速幂,但需要求出一次卷积的效果(每一个位置由所有位置中若干个位置 1 的个数的奇偶性唯一决定),总复杂度 $O(n^4 \log t)$。