2014-2015 ACM-ICPC, Asia Xian Regional Contest C – The Problem Needs 3D Arrays(最大密度子图) 题目大意:给你N个数,要求你找出k个数,使得xn/k的最大值。(xn代表的是k个数组成的逆序对的数量) 解题思路:将每个数当成一个点,如果两个数组成逆序对的关系,就连边,这样图就构成了 详解请看刘伯涛:最小...
本文主要给出了2014-2015 ACM-ICPC, Asia Xian Regional Contest的部分题解,说明了每题的题意、解题思路和代码实现,意即熟悉区域赛比赛题型。 Built with Qinghuai and Ari Factor 题意 判断是否是Q数列,只要数列中每个数均能够被3整除就是Q数列。 解题思路 需要特判一下0的情况。 代码 1#include <cstdio>2...
在刚刚结束的第43届ACM国际大学生程序设计竞赛亚洲区总决赛(Asia-East Continent Final)中,由中山大学数据科学与计算机学院的三名本科生组成的队伍,在郭嵩山老师的带领下,获得了学校亚军,同时获得了2019年参加第43届在葡萄牙举行的国际大学生程序设计竞赛全球总决赛资格。 据悉,该赛事吸引了全国380余支队伍,180多所高校...
#include<bits/stdc++.h>#define int long longusingnamespacestd;#define endl '\n'intmx,my;strings;voidwork(){cin>>mx>>my;cin>>s;intn=(int)s.size();s=' '+s;intnx=0,ny=0;for(inti=1;i<=n;i++){if(s[i]=='U')ny++;if(s[i]=='D')ny--;if(s[i]=='R')nx++;if(s...
代码: intn,k; vector<int>mp[maxn];intnum[maxn];boolisA[maxn];intans=0;voiddfs(intu,intf) {intcnt=0;for(inti=0;i<mp[u].size();i++) {intv=mp[u][i];if(v!=f) { dfs(v,u); num[u]+=num[v];if(num[v]) cnt++; ...
...,n ,满足最长上升序列为 n−1。 贡献为 Ak×(n−k−1)。 代码: int tc = 0; inline void solve() { cout << "Case #" << ++tc << ": "; int n = read(), k = read(), p = read(); if (k >= n - 1) { int ans = 1; for (int i = 1; i <= n; i++)...
The 2018 ACM-ICPC Asia Qingdao Regional Contest, Online H Traveling on the Axis(dp),TravelingontheAxis题解:考虑当前状态和下一个状态的关系,只有四种情况,1→0,1→1,0→1,0→01\to0,1\to1,0\
Key words: ACM/ICPC; problems; Asia regional contest; data; algorithm (编辑:彭远红) 2012-07-13###2012-07-13###2#0#1#2#-07-13### Your request could not be processed because of a configuration error: "Could not connect to LDAP server." For assistance, contact your ...
The 2009 ACM-ICPC Asia Ningbo Regional Online Contest - DDiamonds --- Time Limit: 1 Second Memory Limit: 32768 KB Special Judge --- Task Formally, a diamond with radius r, centered at (x,y), is the set of points whose manhattan distance to (x,y...
The 2009 ACM-ICPC Asia Ningbo Regional Online Contest - DDiamonds --- Time Limit: 1 Second Memory Limit: 32768 KB Special Judge --- Task Formally, a diamond with radius r, centered at (x,y), is the set of points whose manhattan distance to (x,y...