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  • Codeforces 1359D - Yet Another Yet Another Task (最大子段和/枚举)2020-05-29 12:02:05

    https://codeforces.com/contest/1359/problem/D 题意 要求找到“区间和-区间最大值”的最大值 限制 Time limit per test: 1.5 seconds Memory limit per test: 512 megabytes 1≤n≤105 −30≤ai≤30 解题思路 因为答案最小为 0 所以可以从 1 到 30 枚举可能的区间最大值

  • 修改jsp页面内容tomcat报错2020-05-26 10:02:22

    eclipse每次修改jsp页面总会报错“ Publishing failed with multiple errorsCould not delete E:\apache-tomcat-7.0.69\webapps\SafeStorage\logs\all.log. May be locked by another process.Could not delete E:\apache-tomcat-7.0.69\webapps\SafeStorage\logs\debugs.log.

  • CF1324A Yet Another Tetris Problem2020-05-14 16:01:45

                          看起来很高大上的题,其实就是判断这一堆数的奇偶是否一致 过于简单没有代码

  • Yet Another Task with Queens2020-05-05 15:07:18

    Yet Another Task with Queens 思路 我可能写的十分暴力吧,直接四次\(sort\),每一次sort进行一次手动\(unique\)判重。 不难发现每个皇后有八个攻击方向,但是大概可以合并为四大类: \(x\)方向的 \(y\)方向的 斜向上\(45°\)角的 斜向下\(45°\)角的 当我们按照上面的四大类分别排好

  • CF678E Another Sith Tournament(思维+dp)2020-05-04 10:52:15

    题意:自己去翻cf 思路: 一般这种题,我们可以尝试去确定我们dp里放的到底是啥。 在这题里就是结果为1胜利的概率(我们只用一维dp即可; 既然我们是以这个为dp里放的东西,那顺着dp你肯定就算想出来也觉得麻烦, 那考虑倒序,哪两个打完是现在这个状态(https://blog.csdn.net/weixin_43826249/art

  • yum提示Another app is currently holding the yum lock2020-03-20 13:55:07

    使用yum安装计划任务功能,结果提示: yum -y install wgetLoaded plugins: fastestmirror, refresh-packagekit, securityExisting lock /var/run/yum.pid: another copy is running as pid 25960.Another app is currently holding the yum lock; waiting for it to exit...可能是系

  • B - Yet Another Palindrome Problem的简单方法2020-03-17 13:04:10

    You are given an array aa consisting of nn integers. Your task is to determine if aa has some subsequence of length at least 33 that is a palindrome. Recall that an array bb is called a subsequence of the array aa if bb can be obtained by re

  • A. Yet Another Tetris Problem2020-03-16 12:57:44

    A. Yet Another Tetris Problem 题意 你有2X1的矩形,这个矩形只能竖放,不能平放,问最后能不能把题目给出的俄罗斯方块变成一个矩形 。 思路 相邻差不能为奇数。 代码实现 #include<bits/stdc++.h> using namespace std; int main(void){ int t; cin >> t; while(t--){

  • C++派生类的赋值运算符重载详细说明【C++】(zo)2020-03-16 12:05:13

    派生类的赋值符重载情况分析父类和子类都是系统默认子类系统默认,父类自实现赋值运算符重载父类系统默认,子类自实现赋值运算符重载子类和父类都自实现赋值运算符重载父类显式调用语法赋值顺序图示注解代码演示内嵌子对象子类使用系统默认,内嵌子对象类也使用系统默认子类是系

  • [codeforces 1324B] Yet Another Palindrome Problem 回文+边界处理2020-03-13 10:43:23

    Codeforces Round #627 (Div. 3)   比赛人数6434 [codeforces 1324B]  Yet Another Palindrome Problem   回文+边界处理 总目录详见https://blog.csdn.net/mrcrack/article/details/103564004 也在线测评地址http://codeforces.com/contest/1324/problem/B Problem Lang

  • AngularDart初探2020-03-02 09:04:24

    安装pub global activate angular_cli安装完成后需要把pub-cache/bin添加到环境变量中:【windows】添加 C:\Users\Administrator\AppData\Roaming\Pub\Cache\bin 到Path中使用使用ngdart help查看详细命令。Usage: ngdart <command> [arguments]Global options:-h, --help    

  • Yet Another Walking Robot2020-02-05 16:57:37

    题目链接:http://codeforces.com/contest/1296/problem/C 题意:给你一个字符串,是机器人的行走方式,现在要你删除一段连续字符,但机器人最后的终点不变,问你删掉的最短序列是哪一段。           最晚题目意思一直没看懂,以为只能删除前缀或者后缀。 思路:直接用map记录每个状态出现的

  • 自考新教材-p1812020-02-04 14:57:19

    源程序: #include<iostream>using namespace std; class another;class Base{private: float x;public: void print(const another &K);};class Derived:public Base{private: float y;}; class another{private: int aaa;public: another() { aaa=100; } friend void

  • spin_lock为什么要关闭抢占?2020-02-02 10:01:24

    The reason that preemption is disabled on a uniprocessor system is: If not: P1 holds the lock and after a time is scheduled out. Now P2 starts executing and let’s say requests the same spinlock. Since P1 has not released the spinlock, P2 must spin and

  • NumPy Advanced Array Manipulation2020-01-31 20:07:29

    原创转载请注明出处:   ·reshape() In many cases, you can convert an array from one shape to another without copying any data. To do this, pass a tuple indicating the new shape to the reshape array instance method.   A multidimensional array can also be res

  • 计算两点间距离2020-01-31 11:55:33

    输入两点坐标(X1,Y1),(X2,Y2),计算并输出两点间的距离。 Input输入数据有多组,每组占一行,由4个实数组成,分别表示x1,y1,x2,y2,数据之间用空格隔开。Output对于每组输入数据,输出一行,结果保留两位小数。Sample Input 0 0 0 1 0 1 1 0 Sample Output 1.00 1.41 #include<iostream>#include<

  • Waiting for another flutter command to release the startup lock...2020-01-28 23:01:37

      平时我们在开发flutter过程中,在执行flutter packages get命令之后,如果运气不好的,命令没有执行成功的话,我们就会遇到这个错误提示: Waiting for another flutter command to release the startup lock... 1 然后你会发现会发现在任何地方执行flutter命令,都会遇到这个错误: tiny

  • C. Yet Another Broken Keyboard--------思维2020-01-28 15:40:06

    Recently, Norge found a string s=s1s2…sn consisting of n lowercase Latin letters. As an exercise to improve his typing speed, he decided to type all substrings of the string s. Yes, all n(n+1)2 of them! A substring of s is a non-empty string x=s[a…b]=

  • UVA12186 工人的请愿书 Another Crisis 题解2020-01-17 17:06:20

    题目:https://www.luogu.com.cn/problem/UVA12186 DP - 树形DP 设 f[i]f[i]f[i] 为让第 iii 个人签字最少需要签字的工人数量 要让一个人签字,需要至少 T% 的人直接下属签字,我们求出这个 T%并向上取整 对于每个人的子节点,按 fff 值升序排列,并取前 T%人 关键代码 void dfs(int

  • ETL - Extract, Transform, Load2019-12-07 19:00:59

    ETL is short for extract, transform, load, three database functions that are combined into one tool to pull data out of one database and place it into another database. Extract is the process of reading data from a database. In this stage, the data

  • 【2019年8月】OCP 071认证考试最新版本的考试原题-第26题2019-11-11 11:00:08

    Choose three Which three actions can you perfom only with system privileges? A) Truncate a table in another schema. B) Access flat files via a database, which are stored in an operating system directory. C) Log in to a database. D) Query any table in a da

  • CentOS7 yum提示:another app is currently holding the yum lock;waiting for it to exit2019-10-28 15:51:24

    CentOS7 yum安装软件提示:   another app is currently holding the yum lock;waiting for it to exit       yum在锁定状态中。可以通过强制关掉yum进程:   rm -f /var/run/yum.pid   然后就可以使用yum了。   转载: https://blog.csdn.net/u011781521/article/details/52

  • C++等号操作符重载2019-10-27 13:51:45

    在新学操作符重载时最令人头疼的可能就是一些堆溢出的问题了,不过呢,只要一步步的写好new 与 delete。绝对不会有类似的问题。 当时我们编译可以通过,但是运行会出错,因为对象s1与s2进行赋值时,采用浅拷贝,导致对象析构时会对同一块内存空间析构两次。也就是说等号操作符“=”,默认是进行

  • 日常口语五2019-09-02 23:57:50

    Hi, guys, Thanks for joining me for another lesson today. Today, we'll be learning how to think in English, it's so important to learn to think in English. Because when you can think in English,  you can communicate with other English people wit

  • Rust中的函数调用2019-08-19 21:01:55

    注意区别语句和表达式哟。 Rust是一门基于表示式的语言,牢记!!! fn main() { println!("Hello world!"); another_function(5, 6); let y = { let x = 3; x + 1 }; println!("The value of y is: {}", y); let x = plus_one(5); println!("The value of

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