View source for 2018-2019 ACM-ICPC, Asia Dhaka Regional Contest

== 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