Kerberos客户端常用命令包括 kinit, klist, kdestroy, and kpasswd,用户使用这些命令管理自己的 ticket. 此外,每台运行Kerberos的机器应该都配置/etc/krb5.conf,At a minimum, it should define a default_realm setting in [libdefaults]. If you are not using DNS SRV records, it must also contain a [r…
專案中選用大名鼎鼎的 Senparc 微信開發套件 獲取臨時票證處理常式的程式碼 (GetgVXinInfo.ashx) using Senparc.Weixin; using Senparc.Weixin.MP; using Senparc.Weixin.MP.Entities; using Senparc.Weixin.MP.CommonAPIs; using ShouJia.BO; using ShouJia.Facades; using ShouJia.Debugger; using S…
Buy the TicketTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6517 Accepted Submission(s): 2720 Problem DescriptionThe "Harry Potter and the Goblet of Fire" will be on show in the next few day…
Firt thought: an variation to LCS problem - but this one has many tricky detail. I learnt the solution from this link:https://github.com/wangyongliang/Algorithm/blob/master/hackerrank/Strings/Square%20Subsequences/main.cc And here is his code with my…
It is about how to choose btw. BFS and DFS. My init thought was to DFS - TLE\MLE. And its editorial gives a very neat BFS based idea which costs much less memory. https://www.hackerrank.com/challenges/beautiful-path/editorial…
It is marked as a NPC problem. However from the #1 code submission (https://www.hackerrank.com/CharlesOfria), it looks pretty much like a Brutal-Force (or simulation) based solution, mixed with some greedy strategies. To me the other NPC "Queens Revi…
传送门 今天在HackerRank上翻到一道高精度题,于是乎就写了个高精度的模板,说是模板其实就只有乘法而已. Extra long factorials Authored by vatsalchanana on Jun 16 2015 Problem Statement You are given an integer N. Print the factorial of this number. N!=N×(N−1)×(N−2)×⋯×3×2×1 Note: Factorials of N>20…
1289. One Way Ticket Time limit: 1.0 secondMemory limit: 64 MB A crowed of volunteers dressed in the star striped overalls have filled the starport. There starcraft leaves to the thorium mines of Haron. Their job will be hard and dangerous. Will many…
Great learning for me:https://www.hackerrank.com/rest/contests/master/challenges/lucky-numbers/hackers/turuthok/download_solution Basically it is memorized search application. And we follow a discrete strategy: split it into digits and go digit by di…
This is 'Difficult' - I worked out it within 45mins, and unlocked HackerRank Algorithm Level 80 yeah! So the idea is straight forward: 1. sort the input array and calculate partial_sum()2. find the negative\positive boundary with the accumulated give…
A sly knapsack problem in disguise! Thanks to https://github.com/bhajunsingh/programming-challanges/tree/master/hackerrank/algorithms/the-indian-jobLesson learnt: The Italian\Indian job is two-way 01 Knapsack. And some complext problem can be convert…
The most interesting, flexible and juicy binary tree problem I have ever seen. I learnt it from here: https://codepair.hackerrank.com/paper/5fIoGg74?b=eyJyb2xlIjoiY2FuZGlkYXRlIiwibmFtZSI6IkJsdWVCaXJkMjI0IiwiZW1haWwiOiJoZWFsdGh5dG9ueUBnbWFpbC5jb20ifQ%…
Something to learn: Rotation ops in AVL tree does not require recursion. https://github.com/andreimaximov/hacker-rank/blob/master/data-structures/tree/self-balancing-tree/main.cpp node *create_node(int val) { node *pNew = new node; pNew->val = val; p…
https://www.hackerrank.com/contests/infinitum-aug14/challenges/jim-beam 学习了线段相交的判断法.首先是叉乘,叉乘的几何意义是有向的平行四边形的面积(除以2就是三角形的面积).如果ABD和ABC正负相反,说明C和D在AB两侧,同样的,再判断A和B是否在CD两侧即可.当某三角形面积为0时,需要判断是否在线段上. #include <iostream> using namespace std; typedef long long…
https://www.hackerrank.com/challenges/service-lane 用RMQ做的,其实暴力也能过~ #include <iostream> #include <vector> #include <cmath> using namespace std; int main() { int n, t; cin >> n >> t; vector<int> vec(n); for (int i = 0; i…
https://www.hackerrank.com/contests/w1/challenges/maximizing-xor/ 找了半天规律,答案竟然是暴力,伤感.我找到的方法是利用规律2^x XOR 2^x - 1会最大,感觉稍微效率高点. int maxXor(int l, int r) { if (l == r) return 0; int p = 1; for (int i = 0; i <= 10; i++) { if (p * 2 > r) break; p *= 2; } i…
https://www.hackerrank.com/contests/w2/challenges/manasa-and-stones 简单题. #include<iostream> using namespace std; int main() { int T; cin >> T; while (T--) { int n, a, b; cin >> n >> a >> b; // a < b if ( a > b) { int tm…
