介绍 http://keplerproject.github.io/orbit/ Orbit是lua语言版本的MVC框架. 此框架完全抛弃CGILUA的脚本模型, 支持的应用, 每个应用可以卸载一个单独的文件中,当然你也可以将它拆为一个文件, 当你需要时候. 此框架运行在WSAPI协议的服务器上,所以可以工作在 Xavante和一些CGI和fastcgi程序上. Orbit is an MVC web framework for Lua. The design is inspired by li
CARDS Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 1448 Accepted: 773 Description Alice and Bob have a set of N cards labelled with numbers 1 ... N (so that no two cards have the same label) and a shuffle machine. We assume that N i
Old Sorting Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu Submit Status Practice LightOJ 1166 Description Given an array containing a permutation of 1 to n, you have to find the minimum number of swaps to sort the array
Cipher Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 20821 Accepted: 5708 Description Bob and Alice started to use a brand-new encoding scheme. Surprisingly it is not a Public Key Cryptosystem, but their encoding and decoding is base
Cow Sorting Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 6664 Accepted: 2602 Description Farmer John's N (1 ≤ N ≤ 10,000) cows are lined up to be milked in the evening. Each cow has a unique "grumpiness" level in the range 1...
题意:给你一个置换P,问是否存在一个置换M,使M^2=P 思路:资料参考 <置换群快速幂运算研究与探讨> https://wenku.baidu.com/view/0bff6b1c6bd97f192279e9fb.html 结论一: 一个长度为 l 的循环 T,l 是 k 的倍数,则 T^k 是 k 个循环的乘积,每个循环分别是循环 T 中下标 i mod k=0,1,2- 的元素按顺序的连接. 结论二:一个长度为 l 的循环 T,gcd(l,k)=1,则 T^k 是一个循环,与循环 T 不一
传送门 流水线上有n个位置,从0到n-1依次编号,一开始0号位置空,其它的位置i上有编号为i的盒子.Lostmonkey要按照以下规则重新排列这些盒子. 规则由5个数描述,q,p,m,d,s,s表示空位的最终位置.首先生成一个序列c,c0=0,ci+1=(ci*q+p) mod m.接下来从第一个盒子开始依次生成每个盒子的最终位置posi,posi=(ci+d*xi+yi) mod n,xi,yi是为了让第i个盒子不与之前的盒子位置相同的由你设定的非负整数,且posi还不能为s.如果有多个xi,