Here are some tips: No.1:Setting goals When you are going to study something new,you should set a goad for yourself.That will make you focus on the objective while you are studying. No.2:Simulation teaching Pretending you are a real teacher,you teach you
题目 给你一幅8*8的棋盘，上面有8个皇后，问最少的移动步数可使8个皇后无法攻击到对方，移动过程中不可以跨越别的皇后。 题解 可以先把所有符合条件的8皇后位置预处理出来（92种），然后用状态压缩去一一匹配。不可跨越这条件其实没有实际意义，因为跨越别的皇后得到的答案一定不会比不跨越
这是一道维护动态中位数的题，与SP15376不同的地方就是它给出的数字是没有顺序的，这一点很需要注意。 维护动态中位数，有两种做法，对顶堆和链表+Hash ，在这里提供对顶堆的做法。 我们维护两个优先队列，一个大根堆和一个小根堆，在读入整个序列时，我们设当前的序列长度为 \(N\)，我们始终要保持
前言 继续ctf的旅程 攻防世界Misc高手进阶区的3分题 本篇是misc_pic_again的writeup 发现攻防世界的题目分数是动态的 就仅以做题时的分数为准了 解题过程 下下来一个png 扔进winhex 无果 扔进stegsolve 发现压缩包 解压得到一个无后缀文件 扔进winhex 得到flag 结语 eas
题目链接 题意分析 怎么说呢 感觉这道题还是找规律套结论 首先 对于一个字符串 我们最直观的想法就是去掉一个字符 然后再在其余n个位置每个位置可以有m-1种字符插入 那么就存在n*(m-1)种方案 但是存在重复的 对于aaabbbccc这种存在一段连续相同字符的字符串 很显然 一段连续相同
Photo: Catherine Falls Commercial/Moment/Getty Images From creating new words like “Kodak,” forcing words together like “Facebook,” or intentionally misspelling phrases like “Krispy Kreme,” companies have a long history of picking phonetically fa
let's make dependency walker fast again原总结优化工具dependency walkerprocess monitorprocess explorer 缘起 最近，经常需要调查 dll 加载失败的问题。前一阵子刚分享了一篇，感兴趣的小伙伴而可以点击 这里。相信，有 windows 开发经验的小伙伴儿一定听过 Dependency Walker
suggest+名词、动名词短语、句子 suggest后面跟的是建议的事情，具有名词的特性。 比如： I suggest this method. I suggest trying once again. I suggest we have dinner now. we have dinner 这个行为说白了还是一件事情。
对应位置的数字之间的关系是不变的，直接利用这一点就可以做开始还写了ida*，结果t飞了 #include<map> #include<queue> #include<time.h> #include<limits.h> #include<cmath> #include<ostream> #include<iterator> #include<set> #include<stack> #includ
An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop
题目链接：https://codeforces.com/contest/630 A - Again Twenty Five! ？ C - Lucky Numbers ？ B - Moore's Law ？ D - Hexagons! ？ J - Divisibility ？ 注意[2,10]的lcm是 \(2^3*3^2*5*7\) ，也就是每种质因子取最高次。为2520。
Problem Description There must be many A + B problems in our HDOJ , now a new one is coming.Give you two hexadecimal integers , your task is to calculate the sum of them,and print it in hexadecimal too.Easy ? AC it ! Input The input contains several test
在CentOS 6.3 x86_64下安装php-mcrypt的时候出现了问题：Error: Cannot retrieve metalink for repository: epel. Please verify its path and try again，需要安装epel源。 解决方法： 一句话：把/etc/yum.repos.d/epel.repo，文件第3行注释去掉，把第四行注释掉。具体如下： 打开/etc/yum.rep
设dp_i为所求答案，每次选择因数的概率相同，设i有x个因数，dp_i=sum(1/x*x_j)+1,(x_j表示第j个因数)，那我们就预处理每个数的因数即可，T=10000，需要预处理出答案 #include<bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) typedef long long LL; typedef pair<int,in
在Scala 中有两种弱变量 val 和 var 他们最主要的区别就是 val不能再赋值，var可以在生命周期中被多次赋值。 val: 1 scala> val msg=”Hello,world!”; 2 msg: java.lang.String = Hello,world! 3 scala> msg=”Hello again,world!”; 4 :5: error: reassignment to val 5
描述 There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2). 输入 Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000). 输出 Print the word “yes” if 3 divide evenly into F(n). Pr
A. Zoning Restrictions Again ou are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an
An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); p
在Scala 中有两种弱变量 val 和 var 他们最主要的区别就是 val不能再赋值，var可以在生命周期中被多次赋值。 val: 1 scala> val msg=”Hello,world!”; 2 msg: java.lang.String = Hello,world! 3 scala> msg=”Hello again,world!”; 4 :5: error: reassignment to val 5 msg=
Discription Do you have spent some time to think and try to solve those unsolved problem after one ACM contest? No? Oh, you must do this when you want to become a “Big Cattle”. Now you will find that this problem is so familiar: The greatest common di
题目描述 There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2). Input Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000). Output Print the word “yes” if 3 divide evenly into F
yum安装时出现：Cannot retrieve metalink for repository: epel. Please verify its path and try again在CentOS 6.3 x86_64下安装php-mcrypt的时候出现了问题：Error: Cannot retrieve metalink for repository: epel. Please verify its path and try again，需要安装epel源。解决
问题：sdk install gradle 2.14.1报错Stop! The archive was corrupt and has been removed! Please try installing again. Found a previously downloaded gradle 2.14.1 archive. Not downloading it again... End-of-central-directory signature not found. Either this
MacBook安装从网页上下载的软件时，点击安装，出现以下错误 解决方法： 1.右击安装程序： 2.按照下面的路径，双击安装程序： 输入电脑密码后，可以进入安装步骤
next([expr]) 概述 取得一个包含匹配的元素集合中每一个元素紧邻的后面同辈元素的元素集合。 这个函数只返回后面那个紧邻的同辈元素，而不是后面所有的同辈元素（可以使用nextAll）。可以用一个可选的表达式进行筛选。直线电机选型 参数 exprStringV1.0 用于筛选的表达式 示例
