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Description: joseph，著名的约瑟夫环问题，学C的都应该知道
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Size: 973 |
Author: liu |
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Description: 这是经典的joseph算法，适用于小型数据的排序，对于大型数据速度较慢
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Size: 218070 |
Author: 王璐 |
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Description: 实现joseph环问题，用visual c++实现，希望对大家有用，谢谢分享
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Size: 7147 |
Author: 熊海鹏 |
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Description: Joseph环的经典算法，里面包括递归和公式算法和详细的讲解。
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Size: 5855 |
Author: 熊万泽 |
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Description: 《网格计算》
【作者】（美）Joshy Joseph, Craig Fellenstein
当您打开电灯，电力网将立即向您传递所需的能量。那么，计算机和网络是不是也可以这样工作呢?答案是肯定的，它们甚至还可以改变您的业务方式。网格计算推动了“按需电子商务”：在任何地方，任何时间，简单或低成本地传递您所需的确切技术资源。您可以利用现有的系统，立即实现所有的功能。不管您是执行人员、策略人员、设计人员、技术人员还是开发人员，本书都将是您的最佳选择。.
本书主要内容：
掌握网格计算和效用计算的底层概念
学习如何开始着手以及从哪些应用开始
了解最新技术和标准..
使用网格计算最大化现有资源的价值
建立灵活性、弹性和可操作性更强的基础设施
基于“感知和响应”提供对数据和资源的即时访问
消除不同的非集成系统的管理负担
在本书中，一流的IBM专家提供了最好的部署实践、集成现有资源的实际指导以及最新的案例分析；而您需要的是利用网格计算的演化来驱动您的商业价值。-"grid computing" [author (U.S.) Joshy Joseph, Craig Fellenstein When you open Electric, the electricity grid will immediately convey to you the necessary energy. Then, the computer and the Internet can not do this kind of work? Answer is yes, They can even change the way your business. Grid computing to promote the "e-business on demand" : at any place, any time, simple low-cost or you need to transfer the exact technical resources. You can use the existing system, the immediate realization of all the functions. Whether you are implementing personnel, strategy, design, technical staff or staff development, the book will be your best choice. . The book main elements : master grid computing and utility of the concept of calculating the bottom started to learn how and what
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Size: 2955264 |
Author: river_side |
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Description: 一个没有错误的约瑟夫环问题的源代码，约瑟夫问题（Joseph）的一种描述方法是：编号为1，2，.......，n的n个人按顺时针方向围坐一个圈，每个人持人一个密码（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。
利用单向循环链表存储结构模拟此过程，按照出列的顺序印出各人的编号
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Size: 1024 |
Author: Li Dongfang |
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Description: joseph，著名的约瑟夫环问题，学C的都应该知道-joseph, the famous Joseph Central issues, the school should know that C
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Size: 1024 |
Author: liu |
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Description: 约瑟夫环的源程序，希望大家能够支持我的多谢大家-Joseph Central source, I hope everyone will support me I would like to thank everyone
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Size: 1024 |
Author: 杰 |
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Description: 这是经典的joseph算法，适用于小型数据的排序，对于大型数据速度较慢-This is the classic joseph algorithm, applicable to small data sorting, slow for large data
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Size: 218112 |
Author: 王璐 |
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Description: 该算法采用顺序表解决经典约瑟夫问题，算法简单易懂，适合初学数据结构认识阅读-The algorithm using the order form to solve the classic Joseph problems, algorithm is simple and easy to understand, understand the data structure suitable for beginners to read
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Size: 1024 |
Author: leihong |
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Description: 实现joseph环问题，用visual c++实现，希望对大家有用，谢谢分享-Joseph ring realize the problem, using visual c++ Realize, in the hope that useful to everybody, thank you share
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Size: 373760 |
Author: |
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Description: 约瑟夫问题求解
波形音频数据传送速率，其值为通道数×每秒数据位数×每样
本的数据位数／8。播放软件利用此值可以估计缓冲区的大小。-Problem Solving Joseph waveform audio data transfer rate, and its value for the channel-per-second data × × median data for each sample the median/8. Player software can be estimated using this value the size of the buffer zone.
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Size: 143360 |
Author: Kurt |
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Description: 用顺序存储得方实现约瑟夫问题，完成输入给定得个数与每次走得步书，并打印出结果-With a side order of storage realize Joseph problems, the completion of input to set the number of step-by-step walk with each book, and print out the results of
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Size: 1024 |
Author: pu |
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Description: Joseph环的经典算法，里面包括递归和公式算法和详细的讲解。-Joseph Central classical algorithm, which includes recursive algorithms and formulas and detailed explanation.
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Size: 5120 |
Author: 熊万泽 |
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Description: Visual C++ 约瑟夫问题的求解,比较形象化-Visual C++ Joseph problem solving, comparative visualization
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Size: 4942848 |
Author: liangmin |
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Description: 1． 以单项循环链表存储结构模拟约瑟夫环问题。即编号为1、2、3…、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。按出列顺序印出各人编号。
2． 演示程序以用户与计算机的对话方式执行，用户输入相应的数据，输出结果显示在其后。
-err
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Size: 1024 |
Author: 小小 |
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Description: 题目：约瑟夫环问题
一．问题描述
设有n个人围做一圈，现从某个人开始报数，数到m的人出列，接着从出列的下一个人开始重新报数，数到m的人又出列，如此下去，直到所有人都出列为止。试设计确定他们的出列次序序列的程序。
-Title: Joseph Central issues 1. Description of the problem has to do lap Wai n individuals, are reported from a few individuals to start, a few people out of the m out, and then a column from the next re-reported the number of individuals, several of the m out of people, so until all that far out. Test designed to determine whether they should set out the order of a sequence of procedures.
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Size: 1024 |
Author: wangshujing |
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Description: 本程序主要是约瑟夫问题。输入为总人数，然后分别输入每人的密码，最后输入第一次报数的上限。输出则是按出列的先后顺序输出编号。
实例
输入为
7
3 1 7 2 4 8 4
6
则对应输出为6 1 4 7 2 3 5-This procedure is the main problem Joseph. The total number of input, and then enter the password for each, and finally enter the number first reported that the upper limit. Output is shown by a sequence of output numbers. Examples of input 731,724,846 for the corresponding output of 6,147,235
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Size: 15360 |
Author: 尹素芳 |
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Description: 数据结构实习，约瑟夫环的C++实现。在VC++6.0下编译成功。-Data structure internships, Joseph Central C++ to achieve. In VC++6.0 to compile the success of the next.
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Size: 1024 |
Author: Jay |
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Description: 约瑟夫问题，四种解法，分别是静态链表，顺序表两种和循环链表-Joseph problems, the four solution, namely a static list, order form and the cycle of two lists
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Size: 3072 |
Author: daiminmin |
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