Description: Josephus问题的解决方法——用数组方法实现的程序源码-solution to the Josephus problem- an source code based on an array of methods used to achieve the program Platform: |
Size: 167936 |
Author:Bill |
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Description: Josephus问题的另一种解决方法
使用循环链表解决的源代码-Josephus problem using an alternative solution to solve the source code of circular linked list Platform: |
Size: 167936 |
Author:Bill |
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Description: 约瑟夫问题的循环链表求解，可以输出每一步的结果-Joseph circular linked list problem solving, can output the results of each step Platform: |
Size: 579584 |
Author:吕正华 |
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Description: 约瑟夫问题，也即猴子选大王的问题。采用循环链表来实现。-Josephus problem, namely the problem of the monkey king election. Achieved using circular linked list. Platform: |
Size: 2048 |
Author:梁运棠 |
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Description: 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。 -The application is 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 and count the man out of the column m his newspaper the next person and from a starting number, count m the man was out of line and so Law repeated until the round table around were all the columns. Platform: |
Size: 13312 |
Author:吕济根 |
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Description: 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) Platform: |
Size: 380928 |
Author:huang |
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Description: Josephus 问题的算法，在一船上有n人，需要救m人，n个人围成一圈，相隔k个人就丢掉一个，直到只剩下m个人-Josephus problem algorithm, there are n people in a boat, people need to save m, n people form a circle, separated by k individuals to lose one, until only the m individual Platform: |
Size: 1024 |
Author:dengxianzhi |
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Description: a solution to josephus problem with a circular link list. a typical examle of circular link list. Platform: |
Size: 1024 |
Author:SAIKAT MONDAL |
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Description: 求解Josephus问题的函数。用整数序列1, 2, 3, ……, n表示顺序围坐在圆桌周围的人，并采用数组表示作为求解过程中使用的数据结构。-Josephus problem solving functions. With a sequence of integers 1, 2, 3, ... ..., n that order around a round table around and use that as an array of data used in the process of solving the structure. Platform: |
Size: 2035712 |
Author:朱朱 |
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Description: 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 Platform: |
Size: 903168 |
Author:枭龙 |
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Description: 数据结构里面基本所有的知识点的代码为全班同学的作业如约瑟夫问题、哈弗曼编码、层次二叉树、深度广度遍历-Data structure inside knowledge of the basic points of all the code for the whole class of operations such as Josephus problem, Ha Fuman coding, hierarchical binary tree, depth, breadth traversal Platform: |
Size: 4884480 |
Author:方式 |
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Description: This Josephus s Problem
it is make a Visual C++
-This is Josephus s Problem
it is make a Visual C++
Platform: |
Size: 99328 |
Author:Starfaith |
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Description: 这是数据结构中的一个问题，用一带头结点的单向链表解决“约瑟夫环”问题。-This is a problem about data structure , using a linked list without the lead node solution " Joseph Ring" problem. Platform: |
Size: 1024 |
Author:ddd |
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Description: 此源代码用于解决Josephus问题的计算，并帮助初学者有效的学习掌握提高C++编程能力-The source code used to solve the problem of computing Josephus, and to help beginners improve effective learning to master programming skills in C++ Platform: |
Size: 1024 |
Author:logionion |
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Description: 约瑟夫环，设有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。以n=8，s=1，m=4为例，若初始的顺序为 n1，n2，n3，n4，n5，n6，n7，n8。则问题的解为n4，n8，n5，n2，n1，n3，n7，n6。-Joseph Wan, with n individuals sitting around a round table, is reported starting from the number of individuals s, m the number of people to the column and the column from the next person to start reporting the number of numbers to m, who has a column, ..., and so forth until all the people all of the column so far. To n = 8, s = 1, m = 4, for example, if the initial order of n1, n2, n3, n4, n5, n6, n7, n8. The solution of the problem n4, n8, n5, n2, n1, n3, n7, n6. Platform: |
Size: 190464 |
Author:ll |
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Description: 约瑟夫问题解决好人与坏人问题：有n个人围成一圈，分别报数，报到为m的出列，也就是经典的约瑟夫问题。先要解决这样一个问题。围成一圈的人有k个好人，k个坏人，在进行约瑟夫问题报数时，符合在坏人没有都出列前第一个好人没有出列这样的情况的m的最小值为多少?-Joseph, good and bad problem-solving problem: there are n people form a circle, respectively, reported that the number of registration for the m out of line, that is the classic Joseph problems. First to solve such a problem. Circle of people have a good k, k a bad man, when the number reported during the Josephus problem, in line with the bad guys do not have a good man out before the first column is not a situation that the minimum number of m? Platform: |
Size: 546816 |
Author:dekayzc |
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Description: 这里有数据结构课程设计的所有题目，有一些题目的参考实验报告,如停车场问题，八皇后问题，约瑟夫环问题-Here are all the data structures course design topics, some topics of the experiment report, such as parking problems, the eight queens problem, Josephus problem Platform: |
Size: 629760 |
Author:li ui lan |
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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 n (n> 0) individuals sitting around a circle clockwise, each person holds a positive integer password. Choose the beginning of a positive integer as the maximum number of reported m, starting clockwise from the first person from a report from the sequence number, m is the stop newspaper to report the number of people reported m out of line, his password as the new m value, the next person from the beginning of his re-reported from a number. This continues until all the columns all up. Let n be the maximum take 30. To design a program to simulate this process, find out the column number sequence. Platform: |
Size: 2048 |
Author:陈玟 |
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