1st, SSH key: Add a pic @ Sep 18 2016 20:26 To note the configuration process on Linux: 2nd,github网站添加key及创建库:             略  3rd,上传至远程库:4th,   clone:…

CrowdFlower Winner's Interview: 1st place, Chenglong Chen The Crowdflower Search Results Relevance competition asked Kagglers to evaluate the accuracy of e-commerce search engines on a scale of 1-4 using a dataset of queries & results. Chenglong Chen…

题目传送门 /* 二分图点染色:这题就是将点分成两个集合就可以了,点染色用dfs做, 剩下的点放到点少的集合里去 官方解答:首先二分图可以分成两类点X和Y, 完全二分图的边数就是|X|*|Y|.我们的目的是max{|X|*|Y|}, 并且|X|+|Y|=n. 修正:实现多连通块染色,然后贪心选择,将两个集合个数差大的连通块优先添加,能尽量使得un*vn最大 */ #include <cstdio> #include <algorithm> #include <cstring&…

题目传送门 /* 官方题解: 这个题看上去是一个贪心, 但是这个贪心显然是错的. 事实上这道题目很简单, 先判断1个是否可以, 然后判断2个是否可以. 之后找到最小的k(k>2), 使得(m-k)mod6=0即可. 证明如下: 3n(n-1)+1=6(n*(n-1)/2)+1, 注意到n*(n-1)/2是三角形数, 任意一个自然数最多只需要3个三角形数即可表示. 枚举需要k个, 那么显然m=6(k个三角形数的和)+k, 由于k≥3, 只要m?k是6的倍数就一定是有解的. 事实上, 打个表应该也能…

http://acm.hdu.edu.cn/showproblem.php?pid=5311 Hidden String Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1499    Accepted Submission(s): 534 Problem Description Today is the 1st anniversar…

How Much Did It Rain? Winner's Interview: 1st place, Devin Anzelmo An early insight into the importance of splitting the data on the number of radar scans in each row helped Devin Anzelmo take first place in the How Much Did It Rain? competition. In…

Facebook IV Winner's Interview: 1st place, Peter Best (aka fakeplastictrees) Peter Best (aka fakeplastictrees) took 1st place in Human or Robot?, our fourth Facebook recruiting competition. Finishing ahead of 984 other data scientists, Peter ignored…

Souvenir  Accepts: 1078  Submissions: 2366  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 262144/262144 K (Java/Others) Problem Description Today is the 1st anniversary of BestCoder. Soda, the contest manager, wants to buy a souvenir for each…

(很久以前做的,现在发一下)最近做了两个CTF,水平太渣,做了没几道题,挑几个自己做的记录一下. mma ctf 1st 之 rps: from socket import * s = socket(AF_INET, SOCK_STREAM) s.connect(('milkyway.chal.mmactf.link',1641)) #s.connect(('127.0.0.1',10001)) print s.recv(1024) payload = 'a'*48 + '\x03\x00\x0…

Today, Soda has learned a sequence whose n-th (n≥1) item is 3n(n−1)+1. Now he wants to know if an integer m can be represented as the sum of some items of that sequence. If possible, what are the minimum items needed? For example, 22=19+1+1+1=7+7+7+1…