2017-2018 Northwestern European Regional Contest (NWERC 2017)

Rank Solved A B C D E F G H I J K
--/-- 6/11 . O . O . . O O O . O

O: 当场通过

?: 赛后通过

.: 尚未通过

A Ascending Photo

unsolved

B Boss Battle

solved by chelly


chelly‘s solution

C Connect the Dots

unsolved

D Dunglish

solved by ABerror


ABerror‘s solution

E English Restaurant

unsolved

F Factor-Free Tree

unsolved

G Glyph Recognition

solved by ABerror


ABerror‘s solution

H High Score

solved by chelly&&ABerror


chelly‘s solution

I Installing Apps

solved by chelly


chelly‘s solution

J Juggling Troupe

unsolved

K Knockout Tournament

solved by ABerror&chelly


chelly‘s solution

Replay

原文地址：https://www.cnblogs.com/Amadeus/p/9861402.html

时间： 2024-07-30 14:00:44

2017-2018 Northwestern European Regional Contest (NWERC 2017)的相关文章

Northwestern European Regional Contest 2017-I题- Installing Apps题解

一.题意 有一个手机,容量为$C$,网上有$N$个app,每个app有个安装包大小$d_i$,有个安装后的占用空间大小$s_i$,安装app是瞬间完成的,即app的占用空间可以瞬间由$d_i$变成$s_i$,而不需要其他多余的空间.问这个手机最多可以安装多少个app,输出最多可以安装的app数量和安装顺序. 二.思路 很显然的$dp$.按照$max(d,s)-s$从大到小排序.$dp[i][j]$表示在前$i$个app中,占用空间不超过$j$的条件下最多可以安装的app的数量.那么,有如下递推式

2017-2018 ACM-ICPC Northern Eurasia (Northeastern European Regional) Contest (NEERC 17)

2017-2018 ACM-ICPC Northern Eurasia (Northeastern European Regional) Contest (NEERC 17) A 题意:有 n 个时刻,第 i 个时刻要么会在 (xi,yi) 生成一个半径为 yi 的圆,要么射击 (xi,yi) 这个点,如果该点在某个圆内则把对应圆删除并输出该圆的标号,否则输出 -1 .任意时刻圆之间不会相交(可以相切). \(n \le 2*10^5, -10^9 \le x_i,y_i \le 10^9, y

2018-2019 ICPC Northwestern European Regional Programming Contest (NWERC 2018)

J题队友犯了初始化的错,白给6发,本来能1A的 B: solver:lzh.czq 就是个拓扑排序 1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ff first 4 #define ss second 5 #define mp make_pair 6 typedef long long ll; 7 typedef pair<int, int> pii; 8 9 vector<int> g[400

2016-2017 ACM-ICPC Northwestern European Regional Programming Contest (NWERC 2016) 个人题解

Problem C Careful Ascent 可怜的我们被卡在了签到题上,当时用的二分来做,结果速度的左右区间写成了[0,1e32],而改成[-1e32,1e32]就通过了,哎-,怎么就没想到去改一下区间的范围呢. 下面是正常的数学解法,类似于解一元一次方程. #include <iostream> using namespace std; typedef long long ll; ll x,y,n; ll a,b;double c; int main(){ cin>>x&g

The 2017 ACM-ICPC Asia Beijing Regional Contest

地址 Rank Solved A B C D E F G H I J 51/384 4/10 . . ? . O O . O . O O: 当场通过 ?: 赛后通过 .: 尚未通过 A Domains unsolved B K-Dimensional Foil unsolved C Graph upsolved by chelly chelly's solution 很显然的思路就是莫队+并查集 但众所周知并查集可以支持可撤销,但不是很好支持可持久化 于是就可以用上回滚莫队的套路了,回滚莫队可以

Gym-101673 :East Central North America Regional Contest (ECNA 2017)(寒假自训第8场)

A .Abstract Art 题意:求多个多边形的面积并. 思路:模板题. #include<bits/stdc++.h> using namespace std; typedef long long ll; const double inf=1e200; const double eps=1e-12; const double pi=4*atan(1.0); int dcmp(double x){ return fabs(x)<eps?0:(x<0?-1:1);} struct

2016-2017 ACM-ICPC Northeastern European Regional Contest (NEERC 16)

D:上下界费用流 将每个点和每个长度D的区间看作边,限制条件看作流量上下界,差分建图,无源汇最大费用费用流,非常巧妙的使用了差分建图. #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #include<queue> #define N 100003 #define inf 2000000000 #defi

2015-2016 ACM-ICPC Northeastern European Regional Contest (NEERC 15)C - Cactus Jubilee

题意:给一颗仙人掌,要求移动一条边,不能放在原处,移动之后还是一颗仙人掌的方案数(仙人掌:无向图,每条边只在一个环中),等价于先删除一条边,然后加一条边 题解:对于一颗仙人掌,分成两种边,1:环边:环上的边2,树边:非环上的边 考虑1.删除树边,那么只需联通两个联通快,那么方案数就是两个联通块乘积-1(除去删除的边) 2.删除环边,那么我们假设删除所有环,那么图变成了深林,方案数就是深林中每棵树任意两点连接,方案数就是全部的和,先维护一个每个环上的点有多少树边,对于每个树边联通块(大小x)共贡献

hdu6206 Apple 2017 ACM/ICPC Asia Regional Qingdao Online

地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=6206 题目: Apple Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 530    Accepted Submission(s): 172 Problem Description Apple is Taotao's favouri