CF313A Ilya and Bank Account 题解

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CF313A Ilya and Bank Account 题解

Update \(\texttt{2021.3.6}\) 经求学的企鹅提醒修改了 Content 部分的数据范围. Content 有一个人的银行账户里有 \(n\) 元钱,他可以删去倒数第二位获最后一位数字来使得账户里面的钱最大.请求出他的银行账户余额的最大值. 数据范围:\(10\leqslant |n|\leqslant 10^9\). Solution 这道题目很简单,枚举倒数第二位和最后一位数字删去之后的账户余额,然后比较哪个更大就好.当然,如果是正数的话不需要删,肯定是越删余额越少的…

[Codeforces Round #186 (Div. 2)] A. Ilya and Bank Account

A. Ilya and Bank Account time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights an…

以bank account 数据为例，认识elasticsearch query 和 filter

Elasticsearch 查询语言(Query DSL)认识(一) 一.基本认识查询子句的行为取决于 query context filter context 也就是执行的是查询(query)还是过滤(filter) query context 描述的是:被搜索的文档和查询子句的匹配程度 filter context 描述的是: 被搜索的文档和查询子句是否匹配一个是匹配程度问题,一个是是否匹配的问题二.实例导入数据 bank account data download 将数据导入到el…

How To Change the Supplier Bank Account Masking in UI (Doc ID 877074.1)

Give Feedback... How To Change the Supplier Bank Account Masking in UI (Doc ID 877074.1) To Bottom In this Document Goal Solution References APPLIES TO: Oracle Payables - Version: 12.1.0<max_ver> and later [Release: 12.1 and later ]…

CF796C Bank Hacking 题解

洛谷链接题目 Although Inzane successfully found his beloved bone, Zane, his owner, has yet to return. To search for Zane, he would need a lot of money, of which he sadly has none. To deal with the problem, he has decided to hack the banks. There are n ban…

Bank Hacking题解

题目: 题意: 有一颗树,你可以断开点(第一个随便断,以后只能是和已经断开的点相临的点),每个点有权值,断开之后,经一条边和两条边可以到达的节点权值加一,问到最后出现过的最大的权值. 分析: 为啥断开之后就剩下的点就不算相邻了呢...(原题说的是掉线,可是掉线之后线依然在啊...),其实题意好像是掉线之后就断开了. 拿这不就简单了,不管从谁开始断,都会分成几个子树,然后子树又只能从根开始断,这样就有,选作根的点权值为原权值,剩下的和它相邻的权值+1,剩下的权值+2. 然后就随便处理一下就好了,为…

Codeforces Round #186 (Div. 2)

A. Ilya and Bank Account 模拟. B. Ilya and Queries 前缀和. C. Ilya and Matrix 考虑每个元素的贡献. 边长为\(2^n\)时,贡献为最大值: 边长为\(2^{n-1}\)时,贡献为前4大值: 边长为\(2^{n-2}\)时,贡献为前16大值: 以此类推. D. Ilya and Roads \(dp(i,j)\)表示前\(i\)个洞修复了\(j\)个的最小代价. \(dp(i,j)=min\{dp(k,j-(i-k))+c\}\)…