资源列表
person
- 声明基类Person，其中有name，age和no，然后分别声明Teacher（教师）类和Cadre（干部）类从Person继承，这两个类派生出新类Teacher_Cadre（教师兼干部）。要求：1、在Teacher类中还包含数据成员title（职称），在Cadre类中还包含数据成员post（职务），在Teacher_Cadre（教师兼干部）中还包含数据成员wages（工资）。2、在派生类Teacher_Cadre的成员函数show中调用Teacher类中的display函数，输出姓名、年龄、职
Compute
- 约瑟夫（Josephus）环问题：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的上限值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人开始重新从1报数，如此下去，直到所有人全部出列为止。 建立n个人的单循环链表存储结构，运行结束后，输出依次出队的人的序号 -Joseph (Josephus) Central issue: numbered 1,2
oil
- 计算起点到终点之间的需停下加油的加油站数，并记录各加油站-Calculated to be between the beginning to the end to stop the number of refueling stations, and record gas stations
point-rectangle
- 点和矩形类的组合实例，结构优化版，输入平面两点坐标值求出矩形的面积、周长等。-Combination of point and rectangle class instances, structural optimization version, enter the two plane rectangular coordinate values calculated area, perimeter and so on.
sanjiao
- 给出三角形的三边a,b,c，求三角形的面积。只有a+b>c,b+c>a,c+a>b时才能构成三角形。设置异常处理，对不符合三角形条件的输出警告信息，不予计算。-Triangle given three sides a, b, c, find the area of the triangle. Only a+ b> c, b+ c> a, c+ a> b only when the triangle. Set exception handling, the co
simpson-cPP
- simpson 求积分估计s impson 求积分估计simpson 求积分估计-simpson simpson quadrature quadrature estimated estimated estimated simpson simpson quadrature quadrature estimates
exp4-1
- 建立来个字符串 ，并且把来个字符串连接起来并输出。-The establishment of a string, and to a string concatenation and output.
hod
- //3. Write a program to find the sum of following series: //1 + 1/2 + 1/3 – 1/4 + 1/5 - …… up to n terms.
OTHER
- 计算采样算法，包括一些具体的操作。还有一些有集的通道设置-Calculate the sampling algorithms, including some specific operation. There are some set of channel settings
Mjtdecoder1
- 为解决信道编码中循环码译码问题 编写了严格尊崇实际电路的（15，7）码梅吉特译码器-In order to solve channel coding middle cyclic codes decoding the problem prepared a strict revered actual circuit' s (15,7) yards Mejit decoder
Pipedet
- this file is about machine vission.
11032-HappyNumber
- A happy number is defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycl