Description: Josephus问题编写程序，对任意输入的n和m，求出最后剩下的人的编号。要求利用线性表保存这n个人，分别用公式化、链接、模拟指针三种描述方法实现。-The Josephus problem program, for arbitrary input of n and m, obtaining the number of the last remaining people. Require the use of linear table to save n individuals, respectively, with formulaic link three analog pointer description method. Platform: |
Size: 3072 |
Author:张蕊 |
Hits:
Description: 基于C语言实现约瑟夫环的问题，运用了链表插入的方法。-Josephus problem based on the C language, the use of a linked list insertion method. Platform: |
Size: 1024 |
Author:李大夫 |
Hits:
Description: 哈夫曼树，用哈夫曼树解决约瑟夫环问题
代码具有良好封装性-Huffman tree Huffman tree the Josephus problem code with good package Platform: |
Size: 2048 |
Author:朱春明 |
Hits:
Description: 约瑟夫环问题求解
约瑟夫环问题的具体描述是：设有编号为1，2，……，n的n(n>0)个人围成一个圈，从第1个人开始报数，报到m时停止报数，报m的人出圈，再从他的下一个人起重新报数，报到m时停止报数，报m的出圈，……，如此下去，直到所有人全部出圈为止。当任意给定n和m后，设计算法求n个人出圈的次序。-Josephus Josephus problem solving specific description is: has numbered 1, 2, ...... n n (n> 0) individuals in a circle, from the start of the first personal number off, stop newspaper report m The number of reported m, a circle, and then re-reported number from the next person from his report m stop the reported number reported m circle out ...... and they continue to do so until all the circle until everyone. When any given n and m, design algorithms to find the order of n individuals a circle. Platform: |
Size: 1024 |
Author:冯读庆 |
Hits:
Description: 用C程序解决约瑟夫斯问题，是一个出现在计算机科学和数学中的问题。在计算机编程的算法中，类似问题又称为约瑟夫环。-In computer science and mathematics, the Josephus Problem (or Josephus permutation) is a theoretical problem related to a certain counting-out game.
Platform: |
Size: 99328 |
Author:王旺 |
Hits:
Description: 约瑟夫环问题的程序清单-Josephus list of procedures-Josephus problem program list-Josephus list of procedures Platform: |
Size: 1024 |
Author:fghfghrtr |
Hits:
Description: 约瑟夫问题：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。-Joseph: Number as 1, 2, ......, n n individuals in a clockwise direction around a circle, each holding a password (positive integer). The beginning optionally a positive integer as the number off the upper limit m, starting from the first person in a clockwise direction starting from a sequence number of newspaper report m the number of stop message. The reported m people out of the line, his password as the new value of m, starting from the next person in a clockwise direction from a number of reported again, and so on until all all out of the line so far. Design of a test procedure to derive the column order. Platform: |
Size: 5120 |
Author:陈霞 |
Hits:
Description: 题目描述：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，如此反复直到所有人全部出列为止。最终显示出列人的队列。
这个问题是约瑟夫问题的实例化，运用了循环链表，详细算法描述及测试结果见Readme。-Subject description: has n individuals sitting around a round table, now from the first s personal Countin, number m out of the line, and then re-start at the next person from the column number, the number to the m out of the line, and so forth until all all out of the line so far. Finally showing as people queue. This problem is Joseph instance, the use of a circular linked list algorithm is described in detail and test results to see Readme. Platform: |
Size: 200704 |
Author:谢澜 |
Hits:
Description: 任务：编号是1，2，……,n的n个人按照顺时针方向围坐一圈，每个人只有一个密码（正整数）。一开始任选一个正整数作为报数上限值m,从第一个仍开始顺时针方向自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。设计一个程序来求出出列顺序。
要求：利用单向循环链表存储结构模拟此过程，按照出列的顺序输出各个人的编号。
测试数据：
m的初值为20，n=7 ,7个人的密码依次为3，1，7，2，4，7，4，首先m=6,则正确的输出是什么？
要求：
输入数据：建立输入处理输入数据，输入m的初值，n ，输入每个人的密码，建立单循环链表。
输出形式：建立一个输出函数，将正确的输出序列
-Tasks: number is 1, 2, ......, n of n people sitting clockwise circle, each person only a password (a positive integer). The beginning optionally a positive integer as the number off the upper limit m still start from the first clockwise order of the number of reported since 1 report m the number of stop message. The reported m people out of the line, his password as the new value of m, starting from the next person in a clockwise direction from a number of reported again, and so on, until all the columns so far. Design a program to derive the column order.
Requirements: one-way circular linked list storage structure to simulate this process, the number of individuals in accordance with the order of the columns output.
Test data:
the initial value of 20 m, n = 7, 7 personal password followed by 3,1,7,2,4,7,4, first of all m = 6, then what is the correct output?
Requirements:
Input data: input processing input data, enter the initial value of m, n, enter a password for each person, Platform: |
Size: 121856 |
Author:李梅 |
Hits:
Description: 约瑟夫问题是个有名的问题：N个人围成一圈，从第一个开始报数，第M个将被杀掉，最后剩下一个，其余人都将被杀掉。例如N=6，M=5，被杀掉的人的序号为5，4，6，2，3。最后剩下1号。
假定在圈子里前K个为好人，后K个为坏人，你的任务是确定这样的最少M，使得所有的坏人在第一个好人之前被杀掉-Josephus problem is a well-known problem: N personal circle, starting from the first report the number of M will be killed first, the last remaining one, the rest will be killed. For example, N = 6, M = 5, the number of people killed 5,4,6,2,3. The last remaining No. 1. Assumed that the first K in circles as a good man, after the K was bad person, your task is to determine this minimum M, makes all the bad guys in a good man to be killed before Platform: |
Size: 1024 |
Author:jamis |
Hits:
Description: 约瑟夫问题
这是17世纪的法国数学家加斯帕在《数目的游戏问题》中讲的一个故事：15个教徒和15 个非教徒在深海上遇险，必须将一半的人投入海中，其余的人才能幸免于难，于是想了一个办法：30个人围成一圆圈，从第一个人开始依次报数，每数到第九个人就将他扔入大海，如此循环进行直到仅余15个人为止。问怎样排法，才能使每次投入大海的都是非教徒。
含C语言大报告-Josephus problem which is the 17th-century French mathematician Gaspard in the " number of games Problems" tell a story: 15 Protestants and 15 non-believers on the deep distress, half of the people must be put into the sea, and the rest in order to survive, so the thought of a way: 30 individuals have formed a circle, started from the first turn reported the number of individuals per count to the ninth he will be thrown into the sea, so the loop until you are down to 15 people so far . Q. How permutation in order to make each into the sea are non-believers. The report containing the C language Platform: |
Size: 382976 |
Author:薛文琦 |
Hits:
Description: 用循环链求解约瑟夫问题,gnu gcc compiler in codeblocks-use circle link list to solve the problem of Josephus Platform: |
Size: 220160 |
Author:郑佳丽 |
Hits:
Description: 约瑟夫环，经典的数据结构问题，适合初学者研究，用C#语言-Josephus, the classic problem of data structures, suitable for beginners, using C# language Platform: |
Size: 25600 |
Author:林宏声 |
Hits:
Description: 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. Platform: |
Size: 4096 |
Author:安然 |
Hits:
Description: 用java数组解决约瑟夫问题：N个人围成一圈，从第一个开始报数，第M个将退出改圈，最后剩下一个，其余人都将被退出。-Josephus problem solving with java array: N personal circle, starting from the first report the number of M-th will quit changing circle, the last remaining one, the rest will be exited. Platform: |
Size: 1024 |
Author:彭凯 |
Hits:
Description: 通过循环链表解决约瑟夫环问题：一群人围坐成环，指定某人报出某个数，并从这个人开始报数，报到制定的数的人站出来，然后下一个人继续报数，依次循环，最后程序显示站出的顺序-Josephus through circular linked list to solve the problem: a group of people sitting in a ring, someone reported a designated number, and this person reported the number of people who report the number of established stand out, and then the next person to continue to report the number, followed by cycles, the program display station in the order Platform: |
Size: 263168 |
Author:Martin Chou |
Hits:
Description: 约瑟夫斯问题求解（约瑟夫斯（Josephus）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，从他在顺时针方向下一个人开始重新从1报数，如此下去，直至所有的人全部出列为止。试设计一个程序，按出列顺序印出各人编号。）本程序就是解决上述问题-Josephus Problem Solving (Josephus (Josephus) a description of the problem is: numbered 1,2, ..., n, n individuals in a clockwise direction around a circle, each holding a password (positive integer). A start optionally a positive integer as the upper limit of the number of reported m, started from the first in a clockwise direction starting from the number reported, report stops at several m. reported m people out of the line, his password as the new m values , in a clockwise direction from the next person he began to re-number from a newspaper, and so on, until all the people all the columns so far. tried to design a program, according to the column order prints each number.) this program is to solve the above problems Platform: |
Size: 1024 |
Author:王贺 |
Hits:
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 (with numbers 1,2,3 ... n, respectively) sitting around a round table. From the number of people reported a few k, m is the number of out of that person his next and from one individual to start off, count to m that another person out so the law of repeated until the round table were all out of the column. Platform: |
Size: 3677184 |
Author:乔乔 |
Hits:
Search - Josephus problem