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测试通过的环境

  • deepin15.11 amd64(Linux Kernel v4.15.0)
  • deepin20beta amd64(Linux Kernel v5.3.0)

定义

包含签名信息的新Section

  • 名称:.signature

  • 类型:

    • 基于公私钥的签名:SHT_SIG_PKEY = 0x80736967

      0x80736967的含义为(0x80 << 24)|('s' << 16)|('i' << 8)|'g'

    • 基于证书的签名:SHT_SIG_CERT = (SHT_SIG_PKEY + 1)

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题目描述

如题,已知一棵包含N个结点的树(连通且无环),每个节点上包含一个数值,需要支持以下操作:

操作1: 格式: 1 x y z 表示将树从x到y结点最短路径上所有节点的值都加上z

操作2: 格式: 2 x y 表示求树从x到y结点最短路径上所有节点的值之和

操作3: 格式: 3 x z 表示将以x为根节点的子树内所有节点值都加上z

操作4: 格式: 4 x 表示求以x为根节点的子树内所有节点值之和

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题目描述

Your task is, given an integer N, to make a palidrome (word that reads the same when you reverse it) of length at least N. Any palindrome will do. Easy, isn’t it? That’s what you thought before you passed it on to your inexperienced team-mate. When the contest is almost over, you find out that, that problem still isn’t solved. The problem with the code is that the strings generated are often not palindromic. There’s not enough time to start again from scratch or to debug his messy code. Seeing that the situation is desperate, you decide to simply write some additional code that takes the output and adds just enough extra characters to it to make it a palindrome and hope for the best. Your solution should take as its input a string and produce the smallest palindrome that can be formed by adding zero or more characters at its end.

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题目描述

As a sponsor of programming contests, Yu has many factors to consider. Recently, he has found that the difficulties of problems can be a serious factor.
For novices, they may simply ignore the problems that are too hard; and for experts in programming contests, an easy problem almost means nothing. Moreover, if a contest consists of both easy and hard questions, it will not satisfy them — but make them unhappy for both reasons.
Therefore, Yu has come up with an idea: holding different kinds of contests! Novices tend to participant in a contest is known as an easy one, but will not join a hard one.

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题目描述

Oldjang has a sequence $A$ of length $n$, the $i_{th}$ number in which is $A_i$. He defined a function $f(l,r) = a_l \oplus a_{l+1} \oplus \cdots \oplus a_r$. It is simple for the cleverest boy oldjang to calculate the function, so he wants to make the problem more difficult for fun. He will perform two operations:

  1. The operation expressed as “$0 \ x \ y$”, which means he changes the $x_{th}$ number to $y$;

  2. The operation expressed as “$1 \ l \ r$”, which means he wants to calculate the function $F(l,r)$. $F(l,r) = f(l,l) \oplus f(l,l +1) \oplus \cdots \oplus f(l,r)\oplus f(l + 1,l +1) \oplus \cdots f(l + 1,r) \oplus \cdots \oplus f(r,r)$.That means $F(l,r)$ equals to the x or sum of all $f(i,j)$ satisfied $l \le i \le j \le r$.

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环境一览

  • Apache Maven
  • IntelliJ IDEA

安装 Apache Maven

  • 确保已配置好 JAVA 环境,命令行输入 echo %JAVA_HOME% 检查是否有正确输出
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