International Collegiate Programming Contest, Egyptian Collegiate Programming Contest (ECPC 2018)
International Collegiate Programming Contest, Egyptian Collegiate Programming Contest (ECPC 2018)
D. The Millennium Prize Problems
求有公约数x的所有数对之积,那么最大公约数为x的乘积只需要约数有x减掉最大公约数为2x,3x的数对之积。
E. Count Permutations
dp[i][j][k]dp[i][j][k]dp[i][j][k]记前i个已经有连续上升长度为j最后一个在后面的数里是第k小。
G. Robots race
LCT维护一个满足条件的最小生成树,加入新边时与树上链比较,判断是否可以替换,如果可以替换,cut&link。
貌似有更容易的方法。
注意到不同权值的边相互之间是独立的,只要判断相同权值的边能否同时在最小生成树上存在即可。维护权值小于xxx的边加入进来之后图的联通状态(可持久化并查集)。
J. The test cases
倒着看,每次可以把xxx变成x2\frac{x}{2}2x或者x2+N2\frac{x}{2}+\frac{N}{2}2x+2N,要从000开始
遍历[0,n−1][0,n-1][0,n−1],最后在1结束。那么不妨把xxx换成n−xn-xn−x,xxx可以变成⌈x2⌉\lceil \frac{x}{2}\rceil⌈2x⌉或⌈x2⌉+N2\lceil \frac{x}{2} \rceil + \frac{N}{2}⌈2x⌉+2N 。x从0开始遍历,在n−1n-1n−1结束。那么就可以贪心了。
每次让x尽量小即可。
L. Reflection
线段树
M. The business man
线段树
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