Description: 约瑟夫（Joeph）问题的一种描述是：编号为1,2,…,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时停止报数。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个人开始重新从1报数，如此下去，直至所有人全部出列为止。试设计一个程序求出出列顺序。-C language linked list data agencies to achieve Josephus problem. n individuals form a circle, each person has a few, began to report the number, the number to that person out and began to report the number to his number. . . Platform: |
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Description: 约瑟夫环是一个数学的应用问题：已知n个人（以编号1，2，3...n分别表示）围坐在一张圆桌周围。从编号为k的人开始报数，数到m的那个人出列；他的下一个人又从1开始报数，数到m的那个人又出列；依此规律重复下去，直到圆桌周围的人全部出列。
-Josephus is the application of a mathematical problem: Given n individuals (in numbers 1,2,3 ... n, respectively) sitting around a round table around. From the number of people began to report the number of k, the number to m the man out of the column his newspaper the next person and from a starting number, the number to the man and the m columns so regular repeated until the round table around were all out of line. Platform: |
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Author:yyy |
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Description: 约瑟夫环问题：设编号为1,2,3,……,n的n(n>0)个人按顺时针方向围坐一圈,每个人持有一个正整数密码。开始时任选一个正整数做为报数上限m,从第一个人开始顺时针方向自1起顺序报数,报到m是停止报数,报m的人出列,将他的密码作为新的m值,从他的下一个人开始重新从1报数。如此下去,直到所有人全部出列为止。令n最大值取30。要求设计一个程序模拟此过程,求出出列编号序列。-Josephus problem: Let numbered 1,2,3, ... ..., n of the n (n> 0) individuals sitting around a circle clockwise, each person has a positive integer password. Choose the beginning of a positive integer as the reported maximum number of m, starting clockwise from the first person from the sequence reported from a number of m is to stop the newspaper to report the number of people reported m the column, his password as a new m value, the next person from the beginning of his re-reported from a number. It goes on, until all the columns up for all. Let n be the maximum take 30. To design a program to simulate this process, find out the column number sequence. Platform: |
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Author:金庸 |
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Description: 基础问题，用循环链表实现约瑟夫问题，数据结构作业-Basic problem, the problem with the circular linked list implementation Joseph, the data structure operations Platform: |
Size: 1024 |
Author:王子剑 |
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Description: 约瑟夫环问题，在C语言开发案例中有介绍 个人有相对简便的方法。-Josephus problem, in the C language development are described in individual cases have a relatively easy way. Platform: |
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Author:占吉翔 |
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Description: 有n个人围坐在一个圆桌周围，现从第s个人开始报数，数到第m的人出列，然后从出列的下一个人重新开始报数，数到第m的人又出列，…，如此反复直到所有的人全部出列为止.利用Josephus环解决此问题。-There are n personal sitting around a round table all around, from the first s personal beginning now number off, to the number of m row, then stood down from the one start number off, to the number of people and m row,... And so on and so on until all the people stood all so far. Use Josephus ring this problem.
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Size: 168960 |
Author:艳 |
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Description: 解决经典问题约瑟夫环的C程序,简单易行，可供C语言和数据结构初学者使用-A C program can be used to Solve the problem of the Joseph ring
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Author:韦航 |
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Description: 约瑟夫（Joseph）问题的一种描述是：编号为1，2，…，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）。开始任选一个正整数作为报数上限值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时停止报数。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个人开始重新从1报数，如此下去，直至所有人全部出列为止。试设计一个程序求出出列顺序。-Joseph (Joseph) a description of the problem is: No. 1, 2, ..., n-n individuals sitting around a circle in a clockwise direction, the person holding a code (positive integer). Start Choose a positive integer m, reported that the number of upper limit from the first person clockwise from 1 order report number, report m stop the reported number. Reported m the person out of the line, his password as the new values of m, starting from the next person in the clockwise direction from a number off again, and so on, until all all columns so far. Try to design a program to calculate the column order. Platform: |
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Author:哈达 |
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Description: 经典的约瑟夫问题，约瑟夫问题c源码，主要是约瑟夫代码实现。对于c的初学者来说还不错。-The classic problem of Joseph, Joseph problems c source code, source implementation of Joseph. Also good for the beginner of c. Platform: |
Size: 1024 |
Author:祝昊 |
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Description: 约瑟夫环队的程序，其中包含四种不同的程序，包括带头结点，不带头结点的队列程序-Josephus team program, which includes four different programs, including to take the lead node does not take the lead in the queue of the node program Platform: |
Size: 11264 |
Author:周琳 |
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Description: [转发]约瑟夫环问题的不带头结点的循环链表解法，可以成功运行哒~-[Forwarded] Josephus problem does not take the lead in the solution of the circular linked list of nodes, you can successfully run da ~ Platform: |
Size: 3072 |
Author:贾世灵 |
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Description: [转发]约瑟夫环问题的带头节点的循环链表的解法-[Forwarded] the solution of the circular linked list of the leading node of the Josephus problem Platform: |
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Author:贾世灵 |
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Description: 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
对C语言的学习非常有用。-Tower of Hanoi
Fibonacci row
Pascal triangle
Three-color chess
Mice have gone astray Officer (a)
Mice have gone astray Officer (2)
Knights go board
Eight Queen
Eight of silver
Game of Life
String to check
Color, tri-color Towers of Hanoi
Knapsack problem (Knapsack Problem)
The number of operator
Monte Carlo method for PI
Eratosthenes screening and quality
Long integer arithmetic (large numbers operation)
Long PI
The greatest common divisor, least common multiple, factoring
Perfect number
Armstrong number
The maximum number of visitors
Sequence type (pre-order type) in sequence rotary
After the order type of computing
On gambling
Wash cards (random number order)
Craps gambling game
The problem of Joseph (Josephus Problem)
Collection of problems
Permutations and combinations
Gray code (Gray Code)
Produce may be the collection of
m elements of a collection of n elements in a subset of
Digital dismantling
Sort
Score Ranking
Select, insert, bubble sort
S Platform: |
Size: 64512 |
Author:李艳文 |
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Description: 约瑟夫环问题的c++解决方法，包括双向链表和单向链表两种解决方法。-Josephus problem c++ solutions, including the doubly linked list, singly linked list are two workarounds. Platform: |
Size: 5258240 |
Author:沈宙 |
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Description: 分别以数组指针和模拟指针的方法解决约瑟夫问题-Respectively pointer array pointer and simulation methods to solve the problem of Joseph Platform: |
Size: 246784 |
Author:张超 |
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Description: 约瑟夫环是一个数学的应用问题：已知n个人（以编号1，2，3...n分别表示）围坐在一张圆桌周围。从编号为k的人开始报数，数到m的那个人出列；他的下一个人又从1开始报数，数到m的那个人又出列；依此规律重复下去，直到圆桌周围的人全部出列。-Josephus is the application of a mathematical problem: Given n individuals (numbered 1, 2, 3, ... n, respectively) sitting around a round table around. From number k people began to count off the number to m the man out of the line his next person and from a number off, the man was out of the line number to m so regularly repeated down until roundtable around of the people all of the columns. Platform: |
Size: 328704 |
Author:张三 |
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Description: 约瑟夫环...一个链表运用的经典案例,自己写的..还算简洁,希望大家给与指导-Joseph ring ... a classic case of linked lists use to write their own .. pretty simple, I hope you give guidance Platform: |
Size: 578560 |
Author:lareina |
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