AtCoder Beginner Contest 137 昨天还是前天写的,忘了,今天补一下 https://atcoder.jp/contests/abc137 A - +-x #include <bits/stdc++.h> using namespace std; int main () { int a, b; cin >> a >> b; cout << max ({a + b, a - b, a * b}); } B - One Clue #include <...
Atcoder ABC137D:Summer Vacation(贪心) D - Summer Vacation#Time Limit: 2 sec / Memory Limit: 1024 MBScore : 400 pointsProblem Statement# There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of Bi after Ai days from the day you ...
We will hold AtCoder Beginner Contest 137. Contest URL: https://atcoder.jp/contests/abc137 Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20190810T2100&p1=248 Duration: TBD (around 2 hours) Number of Tasks: 6 writer:potetisensei HIR180 Drafear yokozuna57 DEGwer ...
AtCoder Beginner Contest 137 D - Summer Vacation———贪心,优先队列,abc137_D-SummerVacationProblemStatementThereareNNNone-offjobsavailable.Ifyoutakethei-thjobandcompleteit,youwillearntherewardofBiB_
abc105 abc106 abc107 abc108 abc109 abc110 abc111 abc112 abc113 abc114 abc115 abc116 abc117 abc118 abc119 abc120 abc121 abc122 abc123 abc124 abc125 abc126 abc127 abc128 abc129 abc130 abc131 abc132 abc133 abc134 abc135 abc136 abc137 A.cpp B.cpp C.cpp D.cpp abc138 abc139 abc...
AtCoder Beginner Contest 280 (挑战3分钟切ABC,但是AI) https://www.bilibili.com/video/BV1DK411X7wE/ AtCoder Beginner Contest 291 | ABC 吊打萌新珍贵录像 | A~G 题解 https://www.bilibili.com/video/BV17D4y137o6/ 算法专题——Atcoder Beginner Contest309 ...
abc137 E - Coins Respawn Bellman-Ford abc051 D - Candidates of No Shortest Paths Floyd-Warshall joi2008yo F - 船旅 Dijkstra abc035 D - トレジャーハント Dijkstra abc095 C - Half and Half 全探索 abc122 B - ATCoder 全探索 pakencamp-2019-day3 C - カラオケ 全探索 sumitrust201...
AtCoder Beginner Contest 137(A-D) ABC签到,细讲CD题。 A: 意外的是发现了max的新用法,可以的。 B: C: 就是对每个字符串排序,然后再把所有字符串排序,两两相等就加1。这里我用的是vector套string。也可以用无序map,时间都差不多。 下面是大佬写的无序map: D: 说实话,当我知道这道题用优先队列的...
思路:因为 \(k\) 比较小,我们直接跑暴力即可 intmain(){ ios::sync_with_stdio(false),cin.tie(nullptr); ll n;cin>>n; ll ans=0; for(inti=1;i<=n;i++) for(intj=1;j<=n;j++) for(intk=1;k<=n;k++) ans+=__gcd(__gcd(i,j),k); ...
一、题目 https://practice.contest.atcoder.jp/tasks/practice_2 二、分析 这里有三组测试用例。 第一组N = 26, Q = 1000。因为N * N < Q,所以可用冒泡排序法。具体代码参考官方给的Sample Code。 第二组N = 26, Q = 100,由于询问次数只有100次,可用插入排序法处理。每插入一个数据...