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【ACM-ICPC 2018 徐州赛区网络预赛】E. End Fantasy VIX 血辣 (矩阵运算的推广)
Morgana is playing a game called End Fantasy VIX. In this game, characters have nn skills, every skill has its damage. And using skill has special con ...
分类:其他好文   时间:2018-09-10 00:53:52    阅读次数:466
ACM-ICPC 2018 徐州赛区网络预赛 H. Ryuji doesn't want to study (线段树)
Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i]. Unfortunately, ...
分类:其他好文   时间:2018-09-09 23:11:16    阅读次数:289
ACM-ICPC 2018 徐州赛区网络预赛
| Rank | Solved | A | B | C | D | E | F | G | H | I | J | K | | | : : | : : | : : | : : | : : | : : | : : | : : | : : | : : | : : | : : | : : | | 157/ ...
分类:其他好文   时间:2018-09-09 22:41:20    阅读次数:346
ACM-ICPC 2018 徐州赛区网络预赛 H. Ryuji doesn't want to study
262144K Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i]. Unfort ...
分类:其他好文   时间:2018-09-09 22:15:52    阅读次数:252
ACM-ICPC 2018 徐州赛区网络预赛 D. EasyMath
ACM ICPC 2018 徐州赛区网络预赛 D. EasyMath 做法: $$f(m,n) = \sum _{i=1}^{m} \mu(in) = \sum_{i=1}^{m}[gcd(i,n)=1]\mu(i)\mu(n) = \mu(n)\sum_{d|n}\mu(d)f(\frac{m}{ ...
分类:其他好文   时间:2018-09-09 22:14:57    阅读次数:298
ACM-ICPC 2018 徐州赛区网络预赛 J Maze Designer
After the long vacation, the maze designer master has to do his job. A tour company gives him a map which is a rectangle. The map consists of N×M litt ...
分类:其他好文   时间:2018-09-09 19:58:18    阅读次数:276
ACM-ICPC 2018 徐州赛区网络预赛 J. Maze Designer (最大生成树+LCA)
题意:一个N M的矩形,每个格点到其邻近点的边有其权值,需要构建出一个迷宫,使得构建迷宫的边权之和最小,之后Q次查询,每次给出两点坐标,给出两点之间的最短路径 分析:可以把每个格点视作视作图的点,隔开两点的边视作图的边,则构建迷宫可以视作求其生成树,剩余的边就是组成迷宫的墙.因为要花费最小,所以使删 ...
分类:其他好文   时间:2018-09-09 18:15:46    阅读次数:213
ACM-ICPC 2018 I. Characters with Hash
I. Characters with Hash Mur loves hash algorithm, and he sometimes encrypt another one's name, and call him with that encrypted value. For instance, h ...
分类:其他好文   时间:2018-09-09 18:14:52    阅读次数:293
ACM-ICPC 2018 沈阳赛区网络预赛 J. Ka Chang (分块思想)
题目链接:https://nanti.jisuanke.com/t/31451 题意: 给你一颗树,树上各点有初始权值,你有两种操作: 1. 给树中深度为l的点全部+x,(根节点为1,深度为0) 2.求出以x为根的子树权值和 思路: 因为第一个操作是对一整层的树节点+x,那么我们可以很容易的标记每一 ...
分类:其他好文   时间:2018-09-09 15:04:09    阅读次数:167
ACM-ICPC 2018 沈阳赛区网络预赛
J. Ka Chang 昨天学弟写的,感觉题目挺好的,今天写了写。 和以前分块一样,但这题很难想到用分块做。 像是一次操作大量的区间单点修改,区间查询…… #include <bits/stdc++.h> using namespace std; typedef long long ll; cons ...
分类:其他好文   时间:2018-09-09 14:43:43    阅读次数:154
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