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[VC/MFC] Josephus DL : 0: 本文档是（作者：钱能）《C＋＋程序设计教程》系列的部分辅助代码。选题编辑：张朝阳责任编辑：徐培忠、林庆嘉有关代码的具体问题请与作者联系。-This document is (Author: money can)
Date : 2026-01-09 Size : 43kb User : 宋栋
[VC/MFC] classicsinC DL : 0: C语言里面的经典算法，包括有老掉牙河内塔费式数列巴斯卡三角形三色棋老鼠走迷官（一）老鼠走迷官（二）骑士走棋盘八个皇后八枚银币生命游戏字串核对双色、三色河内塔背包问题（Knapsack Problem）数、运算蒙地卡罗法求 PI Eratosthenes筛选求质数超长整数运算（大数运算）长 PI 最大公因数、最小公倍数、因式分解完美数阿姆斯壮数最大访客数中序式转后序式（前序式）后序式的运算关于赌博洗扑克牌（乱数排列） Craps赌博游戏约瑟夫问题（Josephus Problem）集合问题排列组合格雷码（Gray Code）产生可能的集合 m元素集合的n个元素子集数字拆解排序得分排行选择、插入、气泡排序 Shell 排序法 - 改良的插入排序 Shaker 排序法 - 改良的气泡排序 Heap 排序法 - 改良的选择排序快速排序法（一）快速排序法（二）快速排序法（三）合并排序法基数排序法搜寻循序搜寻法（使用卫兵）二分搜寻法（搜寻原则的代表）插补搜寻法费氏搜寻法矩阵稀疏矩阵多维矩阵转一维矩阵上三角、下三角、对称矩阵奇数魔方阵 4N 魔方阵 2(2N+1) 魔方阵 -Obsolete Hanoi Tower Fei-style series Pascal triangle Three-color Chess Mouse walking fans Officer (1) Mouse walking fans Officer (2) Knight walking board 8 Queens 8 coins Game of Life String matching Two-color, three-color Hanoi Tower Knapsack problem (Knapsack Problem) The number of operations Monte Carlo method seeking PI Eratosthenes prime number screened seeking Long integer arithmetic (operations of large numbers) Long PI Greatest common factor, least common multiple, factorization Perfect Number 阿姆斯壮数 Maximum number of visitors Type in order to switch the order of type (pre-order type) Follow-style computing About Gambling Wash cards (random order) Craps Casino Games Joseph problems (Josephus Problem) Set problem Permutations Gray code (Gray Code) Generate the set of possible m elements of the collection a subset of n elements Digital dismantling Sort Score Ranking Selection, insertion, bubble sort Shell
Date : 2026-01-09 Size : 431kb User : neoleo
[VC/MFC] Josephus DL : 0: Josephus问题的描述，还望大家多多指教，谢谢（解题报告）-Josephus description of the problem, but we hope a lot of advice, thank you (problem-solving report)
Date : 2026-01-09 Size : 372kb User : huang
[VC/MFC] josephus DL : 0: josephus 这是一个对josephus问题进行简单研究的小程序，对于初学C++的人，绝对是一个绝佳的选择-josephus josephus This is a simple problem of the small program for the novice C++ people, is definitely a great choice
Date : 2026-01-09 Size : 882kb User : 枭龙
[VC/MFC] Josephus DL : 0: 把约瑟夫问题链表实现 -Joseph realized the problem list
Date : 2026-01-09 Size : 1kb User : 陆泽榕
[VC/MFC] Josephus DL : 0: 利用C++建立链表实行对不同数值约瑟夫环的求解-The establishment of the list using C++ implementation of different numerical solution of Josephus
Date : 2026-01-09 Size : 1.12mb User : zhangwei
[VC/MFC] yuesefuhuan DL : 0: 约瑟夫环问题的源代码 acm必须掌握的算法-Josephus source code acm must master algorithm
Date : 2026-01-09 Size : 13kb User : 李帅
[VC/MFC] 11 DL : 0: 数据结构用链表和数组实现约瑟夫环，经调试无误，可以运行-Data structure with linked lists and arrays Josephus debugging correct, you can run
Date : 2026-01-09 Size : 1kb User : 辛泽彦
[VC/MFC] yuesefu DL : 0: 数据结构经典算法约瑟夫问题 c++ 已经测试通过-josephus c++
Date : 2026-01-09 Size : 1kb User : 张永琪
[VC/MFC] yuesefuhuan DL : 0: 约瑟夫环问题，代码简洁，清晰易懂。执行效率很高-Josephus, the code is simple, clear and easy to understand.
Date : 2026-01-09 Size : 1kb User : hello
[VC/MFC] Josephus DL : 0: 约瑟夫环：编号为1,2,3,4…,n的n个人按顺时针方向围坐一圈,每人持有一个密码(正整数).一开始任选一个正整数作为报数的上限值m,从第一个人开始按顺时针方向自1开始顺序报数,报到m时停止.报m的人出列，将他的密码作为新的m值,从他在顺时针方向上的下一人开始重新从1报数,如此下去,直到所有人全部-Josephus code
Date : 2026-01-09 Size : 1kb User : 咕咚
[VC/MFC] cPPJoseph-ring DL : 0: 1. 熟悉线性表的操作和应用； 2. 利用顺序表的存储结构解决约瑟夫环问题。方法实现：利用线性表中循环链表的结构特点，每个节点表示一个人。用一个程序模拟人的报数出圈，1表示人还在圈里，0表示人已经出圈了。当第L个人出圈之后，报数程序结束，输出最后的在圈子里的人的情况。-A familiar linear form of the operation and application 2. Use sequence table storage structure to solve Josephus problem. Ways: using the linear table circular list of structural features, each node represents a person. With a program to simulate the number of people reported a circle, a circle indicates people are still, 0 means people have been out of the ring. When the first person out of line L, the number of procedures at the end of the final output in the circle of his condition.
Date : 2026-01-09 Size : 4kb User : 安然