【CTF】Reverse [email protected]

来源: Cyber Security Challenge 2015 题目:Crackme 类型: Reverse 分数: 120 难度:中(难中易三个级别) 描述: We found this binary, but we lost the password. Please retrieve it for us. 题目链接:https://github.com/ctfs/write-ups-2015/tree/master/cyber-security-challenge-2015/revers

来源: Backdoor CTF 2015 Little Suzie started learning C. She created a simple program that echo's back whatever you input. Here is the binary file. The vampire came across this service on the internet. nc hack.bckdr.in 8002. Reports say he found a flag

问题:给定一个字符串,字符串中包含若干单词,每个单词间由空格分隔,将单词逆置,即第一个单词成为最后一个单词,一次类推. 说明:字符串本身可能包含前导空格或后导空格,单词间可能包含多个空格,要求结果中去掉前导和后导空格,单词间空格只保留一个. 与rotate函数类似,先逆置每个单词,再将所有字符串逆置. void reverseWords(string &s) { if(s.size() == 0) return; char blank = ' '; size_t len = s.size();

陆陆续续几个月下来,终于把题刷完了,过程中遇到的python的题解很少,这里重新用python实现下,所以题解可能都是总结性的,或者是新的心得,不会仅针对题目本身说的太详细. def reverseWords(self, s): s = ' '.join(s.split()[::-1]) return s [ : :  -1 ] 是将元素进行翻转 [leetcode]Reverse Words in a String (python),布布扣,bubuko.com

Reverse a linked list from position m to n. Do it in-place and in one-pass. For example:Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note:Given m, n satisfy the following condition:1 ≤ m ≤ n ≤ lengt

问题: 给定一个链表的头指针,以及一个整数k,要求将链表按每k个为一组,组内进行链表逆置.少于k个的部分不做处理. 分析: 个人觉得问题的重点是熟悉链表的就地逆置操作,就是头插法.其他的考察点如果还有的话,就的细心程度. 实现: void reverseList(ListNode *&pre, ListNode *head) { ListNode *tail = NULL; while (head) { ListNode* next = head->next; head->next =

Reverse Integer Reverse digits of an integer. Example1: x = 123, return 321Example2: x = -123, return -321 click to show spoilers. Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if you have alread

[题目] Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n

[Title]Pico CTF 2014 Write-up snippet [Description] We found this program snippet.txt, but we're having some trouble figuring it out. What's the value of %eax when the last instruction (the NOP) runs? Hint: You may want to convert the assembly into s