Difference between revisions of "2018-2019 ACM-ICPC, Asia Dhaka Regional Contest"
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Solved by yanghong. 04:34 (+) | Solved by yanghong. 04:34 (+) | ||
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| + | 类似数位dp的想法: 从高位到低位枚举， 如果小于边界则直接计算答案， 等于边界则固定此位置的值然后去低位计算 | ||
== Problem C == | == Problem C == | ||
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Solved by yanghong. 01:06 (+) | Solved by yanghong. 01:06 (+) | ||
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| + | 计算树上路径交 | ||
== Problem G == | == Problem G == | ||
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Solved by bingoier. 00:07 (+) | Solved by bingoier. 00:07 (+) | ||
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| + | 签到，将题面上的代码复制过来即可 | ||
Latest revision as of 12:41, 3 November 2020
Problem A
Unsolved.
Problem B
Solved by yanghong. 04:34 (+)
类似数位dp的想法: 从高位到低位枚举， 如果小于边界则直接计算答案， 等于边界则固定此位置的值然后去低位计算
Problem C
Solved by bingoier. 00:35 (+)
质因数分解后发现答案为每一个质数的指数 $k,((k+2)*(k+1)/2$ 的乘积之和
Problem D
Unsolved.
Problem E
Solved by bingoier. 01:17 (+1)
模拟题，将时间转化为秒数可以简便运算
Problem F
Solved by yanghong. 01:06 (+)
计算树上路径交
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 (+)
签到，将题面上的代码复制过来即可