B. Game of Credit Cards time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and deci…

题目链接: http://codeforces.com/problemset/problem/777/B 题目大意: A, B玩游戏,每人一串数字,数字不大于1000,要求每人从第一位开始报出数字,并且比较,如果等于则没事,A>B则B被打一下,反之A被打一下,A很老实不会耍计谋,老老实实从第一个开始报,但是B很狡猾,会改变数字的顺序.问2个次数,一个是B最小被打的次数,二是A最多被打的次数. 思路: 贪心,模拟一下就行了.这里用到了cmp的技巧,直接用字符比较,这样就省去了转换为字符串的步骤.…

After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards. Rules of this game are simple: each player bring his favourite \(…

B. Game of Credit Cards time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and deci…

B. Game of Credit Cards time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and deci…

After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards. Rules of this game are simple: each player bring his favourite n-…

After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards. Rules of this game are simple: each player bring his favourite n-…

[题目链接]:http://codeforces.com/contest/777/problem/B [题意] 等价题意: 两个人都有n个数字, 然后两个人的数字进行比较; 数字小的那个人得到一个嘲讽; 问你如何搞 才能让莫里亚蒂得到的嘲讽最少; 莫里亚蒂得到的嘲讽最多; [题解] /* 先考虑莫里亚蒂得到最小的; 先把夏洛克的从小到大排序 然后从小到大处理夏洛克的数字; 对于夏洛克的每一个数字x 尝试用一个>=x的但是最小的数来击败它.然后这个数打上标记; 如果找不到大于等于它的数字,那么就结…

对第一个人的排序,然后从小到大处理,对第一个人的每枚卡片,从第二个人的卡片中选择一个大于等于它的最小的,否则选择一个当前剩下的最小的,这样可以保证负场最少. 如果选择的改成大于它的最小的,就可以保证胜场最多. 用multiset处理. #include<cstdio> #include<set> #include<algorithm> using namespace std; typedef pair<int,int> Point; Point c[1010…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are different. They divid…

Cards 题意: 给你n个牌,n是偶数,要你把这些牌分给n/2个人,并且让每个人的牌加起来相等. 题解: 这题我做的时候,最先想到的是模拟,之后码了一会,发现有些麻烦,就想别的方法.之后发现只要把它给的数排序就好了,排完序后输出原来的位置就好,因为每2个都能配对所以最小肯定配最大,第二小配第二大-- 之后就可以a了 代码: #include <bits/stdc++.h> using namespace std; const int INF=0x3f3f3f3f; typedef long…

Soldier and Cards 老样子,直接上国语吧  Descriptions: 两个人打牌,从自己的手牌中抽出最上面的一张比较大小,大的一方可以拿对方的手牌以及自己打掉的手牌重新作为自己的牌,放在自己手牌的最下方,而且对方输掉的那张手牌需要放在上面,自己赢的手牌放在下面. Input 第一行的数n代表一共有几张牌 第二行第一个数x代表第一个人有x张牌 第三行第一个数y代表第二个人有y张牌 Output 第一个数代表进行了几轮,第二个数代表谁赢 Examples Input 42 1 32…

[题目链接]:http://codeforces.com/problemset/problem/743/E [题意] 给你n个数字; 这些数字都是1到8范围内的整数; 然后让你从中选出一个最长的子列; 要求这个子列中各个数字出现的次数的差的绝对值都不超过1; 且如果是相同的数字的话: 都是连在一起的(不会有分散的数字); 问你这个最长的序列的长度是多少; [题解] 二分每个数字至少出现的次数x,(即最少出现x次,当然也可以是x+1次); (单调性是显然的吧,因为如果每个数字出现5次是可行的话,那…

Credit Card time limit per test2 seconds memory limit per test256 megabytes Recenlty Luba got a credit card and started to use it. Let's consider n consecutive days Luba uses the card. She starts with 0 money on her account. In the evening of i-th da…

这两天回家了 家里电脑太卡 调试不方便 就只能写写水题了…… #include<stdio.h> #include<iostream> #include<algorithm> #include<math.h> #include<string.h> #include<string> #include<map> #include<set> #include<vector> #include<que…

我只能说真的看不懂题解的做法 我的做法就是线段树维护,毕竟每个数的顺序不变嘛 那么单点维护 区间剩余卡片和最小值 每次知道最小值之后,怎么知道需要修改的位置呢 直接从每种数维护的set找到现在需要修改的数的在初始卡片的位置 #include <cstdio> #include <cstdlib> #include <cstring> #include <ctime> #include <algorithm> #include <iostre…

Codeforces Round #401 (Div. 2) 很happy,现场榜很happy,完全将昨晚的不悦忘了.终判我校一片惨白,小董同学怒怼D\E,离AK就差一个C了,于是我AC了C题还剩35分钟立刻开D题,不料英文苦涩难懂,懂了题意之后发现也是个水题,从后往前遍历一遍即可.然而代码速度和思维都不算很好,还有半分钟在徘徊要不要测一下样例,但又怕OJ卡了,直接交第一组就跪了,然后发现个小问题改改又直接交最后5秒然而以第一组WA结束本场CF. A  Shell Game 题意:有三个位置分别…

http://codeforces.com/contest/701 A - Cards 水 // #pragma comment(linker, "/STACK:102c000000,102c000000") #include <iostream> #include <cstdio> #include <cstring> #include <sstream> #include <string> #include <alg…

A. Shell Game time limit per test 0.5 seconds memory limit per test 256 megabytes input standard input output standard output Bomboslav likes to look out of the window in his room and watch lads outside playing famous shell game. The game is played b…

A. Shell Game time limit per test 0.5 seconds memory limit per test 256 megabytes input standard input output standard output Bomboslav likes to look out of the window in his room and watch lads outside playing famous shell game. The game is played b…

B B - Local Extrema CodeForces - 888A You are given an array a. Some element of this array ai is a local minimum iff it is strictly less than both of its neighbours (that is, ai < ai - 1 and ai < ai + 1). Also the element can be called local maximum…

byl太强了,学弟们太强了-全程被吊打,嘤嘤嘤- A题  Connecting Vertices http://codeforces.com/problemset/problem/888/F 不会 B题 Local Extrema http://codeforces.com/problemset/problem/888/A 给一列数字,判断一个数它的左右是否同时比它大,或者同时比它小,若满足的话那么count++,得到最后的count值,那么很明显,直接暴力的遍历一遍就可以了. 1 #inclu…

原文 We can't move beyond the past until our emotional debts are paid. Many people today live under the weight of loans and credit cards.As harmful as financial debt is, there is another form of debt that is even more damaging-emotional debts. Dr.James…

My name is Charles Humble and I am here at QCon New York 2014 with Ian Robinson. Ian, can you introduce yourself to the InfoQ community? Hello, I am Ian Robinson, I am engineer at Neo Technology, I am based in London and I work on the Neo4j graph dat…

What things should a programmer implementing the technical details of a web application consider before making the site public? If Jeff Atwood can forget about HttpOnly cookies, sitemaps, and cross-site request forgeries all in the same site, what im…

Anomalies are data points that are few and different. As a result of these properties, we show that, anomalies are susceptible to a mechanism called isolation. This paper proposes a method called Isolation Forest (iForest) which detects anomalies pur…

Source: http://www.wantchinatimes.com/news-subclass-cnt.aspx?id=20140510000005&cid=1103 China UnionPay, China's leading issuer of credit cards, has implemented an anti-money laundering campaign in Macau as tens of billions of yuan have been illegally…

Electronic Payment App is getting more and more popular now. People don't have to bring credit cards any more. All they need to do is using their smartphones and they could go shopping, check bills and dining in restaurants. It very convenient but so…

在Chrome中以https访问自己搭建的gitlab站点时经常出现下面的错误: Attackers might be trying to steal your information from xxx.com (for example, passwords, messages, or credit cards). NET::ERR_CERT_AUTHORITY_INVALID xxx.com normally uses encryption to protect your informatio…

RFID读写器的工作原理 RFID的数据采集以读写器为主导,RFID读写器是一种通过无线通信,实现对标签识别和内存数据的读出和写入操作的装置. 读写器又称为阅读器或读头(Reader).查询器(Interrogator).读出装置(Reading Device). 扫描器(Scanner).通信器(Communicator).编程/编码器(Programmer)等等. 读写器工作原理 RFID读写器的基本原理是利用射频信号与空间耦合传输特性,使电子标签与阅读器的耦合元件在射频耦合通道内进行能量传…