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Description: Josephus 排列问题定义如下：假设n 个竞赛者排成一个环形。给定一个正整数m，从某
个指定的第1 个人开始，沿环计数，每遇到第m 个人就让其出列，且计数继续进行下去。这
个过程一直进行到所有的人都出列为止。最后出列者为优胜者。每个人出列的次序定义了整
数1，2，…，n 的一个排列。这个排列称为一个(n,m)Josephus 排列。
例如，(7,3)Josephus 排列为3，6，2，7，5，1，4。-Josephus problem with the definition is as follows : Suppose n race who formed a ring. Given a positive integer m, from a certain section of a designated individual, along the Central Counting that every individual section m let out its out and count continue. This process continues until all the people from far out. Finally out of the winners were shown. Everyone out in the order shown in the definition of integers 1, 2, ..., n an order. With this as a (n, m), with Josephus. For example, (7,3) Josephus, were 3,6,2,7,5,1,4.
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Size: 1095 |
Author: 李飞飞 |
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Description: 片机开发过程中用到的多功能工具，包括热敏电阻RT值--HEX数据转换；3种LED编码；色环电阻计算器；HEX/BIN 文件互相转换；eeprom数据到C/ASM源码转换；CRC校验生成；串口调试，带简单而实用的数据分析功能；串口/并口通讯监视等功能. 用C++ Builder开发，无须安装，直接运行，不对注册表进行操作。纯绿色软件。-machine used in the process of developing the multi-purpose tools, including thermistor RT -- HEX data conversion; 3 LED coding; Color ring resistance calculators; HEX / BIN documents interchangeable; eeprom data to the C / ASM source code conversion; CRC generation; Serial debugging, with a simple and practical data analysis functions; Serial / Parallel Communications surveillance, and other functions. C Builder development, no installation, operation directly, and do not operate the registry. Green pure software.
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Size: 1391078 |
Author: 赵相立 |
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Description: Josephus排列问题定义如下：假设n个竞赛者排成一个环形。给定一个正整数m，从某个指定的第一个人开始，沿环计数，每遇到第m个人就让其出列，且计数继续进行下去。这个过程一直到所有的人都出列为止。最后出列都优胜者。每个人出列的次序定义了整数1，2，...，n的一个排列。这个排列称为一个(n,m)Josephus排列。例如，（7,3)Josephus排列为3,6,2,7,5,1,4.对于给定的1，2，...n中的k个数，Josephus想知道是否存在一个正整数m(n,m)Josephus排列的最后k个数为事先指定的这k个数。-Josephus problem with the definition is as follows : Suppose n player formed a ring. Given a positive integer m, from a designated first started along the Central Counting that every individual section m let out their series and counting continued. This process has been to all of them from far out. Finally out from all the winners. Everyone out in the order shown in the definition of integers 1, 2 ,..., n an order. This arrangement known as a (n, m) Josephus arranged. For example, (7,3) Josephus order of 3, 6,2,7,5,1,4. For a given set of 1,2, ... n and k the number, Josephus would like to know whether there is a positive integer m (n, m) Josephus with the final number k prior to the designated number k this.
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Size: 2011 |
Author: 诛仙 |
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Description: Josephus 排列问题定义如下：假设n 个竞赛者排成一个环形。给定一个正整数m，从某
个指定的第1 个人开始，沿环计数，每遇到第m 个人就让其出列，且计数继续进行下去。这
个过程一直进行到所有的人都出列为止。最后出列者为优胜者。每个人出列的次序定义了整
数1，2，…，n 的一个排列。这个排列称为一个(n,m)Josephus 排列。
例如，(7,3)Josephus 排列为3，6，2，7，5，1，4。-Josephus problem with the definition is as follows : Suppose n race who formed a ring. Given a positive integer m, from a certain section of a designated individual, along the Central Counting that every individual section m let out its out and count continue. This process continues until all the people from far out. Finally out of the winners were shown. Everyone out in the order shown in the definition of integers 1, 2, ..., n an order. With this as a (n, m), with Josephus. For example, (7,3) Josephus, were 3,6,2,7,5,1,4.
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Size: 1024 |
Author: 李飞飞 |
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Description: 片机开发过程中用到的多功能工具，包括热敏电阻RT值--HEX数据转换；3种LED编码；色环电阻计算器；HEX/BIN 文件互相转换；eeprom数据到C/ASM源码转换；CRC校验生成；串口调试，带简单而实用的数据分析功能；串口/并口通讯监视等功能. 用C++ Builder开发，无须安装，直接运行，不对注册表进行操作。纯绿色软件。-machine used in the process of developing the multi-purpose tools, including thermistor RT-- HEX data conversion; 3 LED coding; Color ring resistance calculators; HEX/BIN documents interchangeable; eeprom data to the C/ASM source code conversion; CRC generation; Serial debugging, with a simple and practical data analysis functions; Serial/Parallel Communications surveillance, and other functions. C Builder development, no installation, operation directly, and do not operate the registry. Green pure software.
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Size: 1390592 |
Author: 赵相立 |
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Description: Josephus排列问题定义如下：假设n个竞赛者排成一个环形。给定一个正整数m，从某个指定的第一个人开始，沿环计数，每遇到第m个人就让其出列，且计数继续进行下去。这个过程一直到所有的人都出列为止。最后出列都优胜者。每个人出列的次序定义了整数1，2，...，n的一个排列。这个排列称为一个(n,m)Josephus排列。例如，（7,3)Josephus排列为3,6,2,7,5,1,4.对于给定的1，2，...n中的k个数，Josephus想知道是否存在一个正整数m(n,m)Josephus排列的最后k个数为事先指定的这k个数。-Josephus problem with the definition is as follows : Suppose n player formed a ring. Given a positive integer m, from a designated first started along the Central Counting that every individual section m let out their series and counting continued. This process has been to all of them from far out. Finally out from all the winners. Everyone out in the order shown in the definition of integers 1, 2 ,..., n an order. This arrangement known as a (n, m) Josephus arranged. For example, (7,3) Josephus order of 3, 6,2,7,5,1,4. For a given set of 1,2, ... n and k the number, Josephus would like to know whether there is a positive integer m (n, m) Josephus with the final number k prior to the designated number k this.
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Size: 2048 |
Author: 诛仙 |
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Description: Josephus 排列问题定义如下：假设n 个竞赛者排成一个环形。给定一个正整数m，从某
个指定的第1 个人开始，沿环计数，每遇到第m 个人就让其出列，且计数继续进行下去。这
个过程一直进行到所有的人都出列为止。最后出列者为优胜者。每个人出列的次序定义了整
数1，2，…，n 的一个排列。这个排列称为一个(n,m)Josephus 排列。
例如，(7,3)Josephus 排列为3，6，2，7，5，1，4。
对于给定的1，2，…，n 中的k 个数，Josephus 想知道是否存在一个正整数m 使得
(n,m)Josephus 排列的最后k 个数恰为事先指定的这k 个数。-Josephus with the problem are defined as follows: Suppose n a player lined up a ring. Given a positive integer m, from a designated section 1 begins with individuals along the ring count, the first m individuals, when confronted by their let out, and the count continued. This process until all the listed date. Finally a column for the winners. Each person listed in the order of the definition of the integers 1,2, ..., n an order. This arrangement is called a (n, m) Josephus arranged. For example, (7,3) Josephus arranged for 3,6,2,7,5,1,4. For a given 1,2, ..., n in the k number, Josephus would like to know whether there is a positive integer m makes (n, m) Josephus with the final number of exactly k pre-designated number of this k.
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Size: 1024 |
Author: 杨哲 |
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Description: 问题描述：
新Josephus排列问题定义如下：假设n个人排成一个环形，给定一个正整数m，从第1
个人开始，沿环计数，每遇到第m个人就让其出列，这时计算器m自动加一，且计数继续进
行下去。这个过程一直进行到所有的人都出列为止，最后出列者为赢家。设这n个人的编号
分别为1～n，每次从编号为1的竞赛者开始计数，那么每个人出列的次序定义为整数1～n
的一个排列。这个排列称为一个(n,m)Josephus排列。
例如，(7,3)Josephus 排列为3，7，6，2，4，1，5。
对于给定的1，2，…，n中的k个数，Josephus想知道是否存在一个正整数m使得
(n,m)Josephus排列的前k个数恰为事先指定的这k个数。
编程任务：
(1)用抽象数据类型表设计一个求(n,m)Josephus排列的算法。
(2)试设计一个算法，对于给定的正整数n和1，2，…，n中的k个数。求正整数m，使
(n,m)Josephus排列的前k个数恰为事先指定的这k个数(顺序必须完全一样)。
-Problem Description:
Josephus ordered a new problem definition is as follows: Suppose n individuals arranged in a ring, given a positive integer m, from the first one
Begins with individuals along the ring count, when confronted with the first m individuals let their out of line, then automatically add a calculator m, and the count continued into the
OK go on. This process continues until all of them out of the column until the last person out of the column as a winner. This set up the individual numbers n
Respectively, 1 ~ n, each number is one from the race have begun to count, then the order of each person out of the column is defined as an integer 1 ~ n
An array. This arrangement is called a (n, m) Josephus arranged.
For example, (7,3) Josephus arranged as 3,7,6,2,4,1,5.
For a given 1,2, ..., n in the k-number, Josephus would like to know whether there is a positive integer m makes the
(n, m) Josephus arranged in exactly the number of pre-k this for a pre-specified k number
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Size: 2048 |
Author: atlantis |
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Description: 这是在Ring 3环境下简单的API Hook，实现让任务管理器无法结束进程。-This is the Ring 3 environment, a simple API Hook, to achieve so that Task Manager can not end the process.
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Size: 3300352 |
Author: 景甜 |
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Description: 编写一个控制光标位置和形状的程序，改程序具有以下功能
1：可用光标移动键↑↓←→来移动光标2：当光标已在第0列，则按←键，光标定在上一行的最后一列，若已在屏幕的左上角，则光标不动，且给出响铃，按→键时的边界处理类似。3：当光标在第0行，则按↑键，光标不动，且给出响铃，按↓键时的边界处理类似。4：按Home或End键，则光标移到当前行的行首或行尾.5:若按下数字或字母，则把该字符从当前位置依次显示到屏幕顶（在新位置显示当前字符时，原位置的符号被抹去）。6：按Esc ，结束-The preparation of a control cursor position and shape of the process, change process has the following features
1: The available cursor keys ↑ ↓ ← → to move the cursor 2: When the cursor in the first 0, press ← key, the cursor on the line set at the last one, if it has the upper left corner of the screen, the cursor does not move, and given the ring, press → key when dealing with similar border. 3: When the cursor row 0, press ↑ key, the cursor does not move, and given the ring, press ↓ key when dealing with similar boundaries. 4: Press Home or End key, then move the cursor to the first row the current row or the end of the line .5: If press the number or letter, put the order of the characters from the current location to the screen top (in the new location to display the current character The original location of the sign to be erased). 6: press Esc, the end of
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Description: 约瑟夫环的问题采用的是典型的循环链表的数据结构,就是将一个链表的尾元素指针指向队首元素。
解决问题的核心步骤：
1.建立一个具有n个链结点，无头结点的循环链表；
2.确定第1个报数人的位置；
3.不断地从链表中删除链结点，直到链表为空。
具体过程就是先建一个单向循环链表，用来存储这些人的编码和密码。然后删掉报数为m的人，在删除的同时将这个人的密码和标号输出。直到把这个链表删空为止。这就是我解决这个问题的方案。
-Joseph Ring is a typical problem with circular list data structure is a list of the last element pointer to force the first element. The core problem-solving steps: 1. To establish a link with n-points, without a head of circular list node 2. Identified the first one reported the location of several people 3. Continuously removed from the list link point, until the list is empty. Specific process is to first build a one-way circular linked list, these people used to store code and password. Then delete the number m of people reported in these individuals at the same time remove the password and label output. Delete this list until it is empty. This is what I solve this problem.
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Size: 1024 |
Author: akon_shuai |
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Description: 单片机开发过程中用到的多功能工具，包括热敏电阻RT值--HEX数据转换；3种LED编码；色环电阻计算器；HEX/BIN 文件互相转换；eeprom数据到C/ASM源码转换；CRC校验生成；串口调试，带简单而实用的数据分析功能；串口/并口通讯监视等功能. 用C++ Builder开发，无须安装，直接运行，不对注册表进行操作。纯绿色软件。-Microcontroller used in the process of developing multi-functional tools, including a thermistor RT value- HEX data conversion three kinds of LED codes color ring resistance calculator HEX/BIN files interchangeable eeprom data to the C/ASM source conversion CRC checksum generation serial debugging, with simple and practical data analysis serial/parallel communication monitoring and other functions. using C++ Builder developers, no installation, run directly, not operate the registry. Pure green software.
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Size: 1389568 |
Author: xialu |
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Description: joseph环
*问题描述：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？
*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
-joseph ring* Description: The number is 1, 2, ......, n-n individuals sitting around the circle clockwise, each person has only one password (positive integer). Choose a positive integer as the reported number of the upper limit of m,, since the number of a start sequence reported from the first one is still clockwise, the 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. Design a program to calculate the column order.* Requirements: one-way circular linked list storage structure simulation of this process, in accordance with the order of the columns to output the number of individuals. Test data: initial value of m 20, n = 7, 7 personal password followed by 3,1,7,2,4,7,4, first of all m, = 6, then the correct output is what? Input data: the establishment of input processing the input data,
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Author: 黄敏 |
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Description: 简单的防止任务管理器结束进程~~3环的-Prevent the Task Manager to end the process ~ ~ ring ~ ~
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Size: 86016 |
Author: 闫斌 |
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Description: 约瑟夫环问题
【问题描述】
Josephus排列问题定义如下：假设n个竞赛者排成一个环形。给定一个正整数m≤n，从第1人开始，沿环计数，第m人出列。这个过程一直进行到所有人都出列为止。最后出列者为优胜者。全部出列次序定义了1，2，…n的一个排列。称为（n，m）Josephus排列。例如，（7，3）Josephus排列为3,6,2,7,5,1,4。
【实验要求】
设计求解Josephus排列问题程序。
（1）采用顺序表、单链表或双向循环链表等数据结构。
（2）采用双向循环链表实现Josephus排列问题，且奇数次顺时针轮转，偶数次逆时针轮转。
（3）推荐采用静态链表实现Josephus排列问题
-Josephus Josephus problem [Problem Description] permutation problem is defined as follows: Suppose the n contestants lined up a ring. Given a positive integer m 鈮?n, beginning from the first one, along the ring count, the first man out of the line m. This process continues until everyone is out of the line so far. Finally out of the line by the winner. All the columns defined sequence 1,2, ... n of an order. Called (n, m) Josephus order. For example, (7,3) Josephus arranged 3,6,2,7,5,1,4. Experimental Design [requirements] Josephus permutation problem solving process. (1) using the order form, single chain or two-way circular linked list data structure. (2) using two-way circular linked list implementation Josephus permutation problem, and the odd clockwise rotation, counterclockwise rotation even number of times. (3) recommend the use of a static list to achieve Josephus permutation problem
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Author: 张小红 |
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Description: 51单片机C语言GPRS控制程序。
2011-05-07更新:
1、优化了模块指令解码程序。
2、增加了AT指令执行 “OK”的检测
3、增加了响铃行为“RING”的检测
4、优化了短信控制指令的判断优化
2011-05-08更新
增加网络注册状态检查
增加GPRS网络附着情况检查
2011-5-22
修改TXD IO口为推勉输出
RXD 为高阻输入-51 microcontroller C language GPRS related procedures. 2011-05-07 update: 1, optimize the module instruction decoding process. 2, increasing the AT command execution " OK" testing 3, an increase of Bell behavior " RING" 4 detection to optimize the SMS control instructions to determine optimized 2011-05-08 update adds GPRS network registration status checks increase network attached conditions check 2011-5-22 modify TXD IO port to push Mian RXD output is high impedance input
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Author: kevin |
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Description: 以正整数n，i，k作为输入，其中n，i，k>0，且i<n。假定数1，2，…，n是环形排列的。试编写一个程序，从数i开始，按顺时针方向以k为步长打印数，当打印某个数时，应从环中删去该数，这样的过程一直进行到环空为止。例如，当n=10，i=1，k=3时，得到的输出序列是3，6，9，2，7，1，8，5，10，4。-In a positive integer n, i, k as an input, wherein n, i, k> 0, and i <n. Assuming that the number 1,2, ..., n is an annular array. Try to write a program, starting number i, k clockwise direction increments the number of prints, when printing a number, this number should ring deleted, this process has been carried out to the ring empty. For example, when n = 10, i = 1, k = 3, the output sequence is obtained 3,6,9,2,7,1,8,5,10,4.
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Author: 郑胜达 |
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