资源列表
_ch301.c
- 回文判断,正与反都用相同的字符序列为准.用堆栈实现-Palindrome judge, is working with the anti-all with the same sequence of characters shall prevail. To use the stack to achieve
main
- josephus环的普通算法，使用指针，编译通过-Josephus ring general algorithm, the use of pointer, the compiler through
sram_w_eye_read.sp
- spice netlist with sense amplifier
fee
- 编写程序用静态成员的方法实现对班费的管理，要求定义一个类Student，除了声明一个存放班费的静态成员，还要分别定义一个上交班费的成员函数Contribute（），花费班费的成员函数Spend（）和显示班费的静态成员函数Display（）-Write a program with static member class fee management, require the definition of a class Student, stored in addition to declare
enhancedsafearray
- This is Exam 1 practical of ssd5, code is compiling and i have got 95 for this file, dwonload-This is Exam 1 practical of ssd5, code is compiling and i have got 95 for this file, dwonload
SOM
- 基于matlab的自组织神经网络进行遥感图像分类处理-Matlab based on the self-organizing neural network for remote sensing image classification processing
byteto16
- 能够将byte字符转换为16进制字符，详细功能见源文件注释部分-Byte characters are converted to hexadecimal characters, functions, see the comment section of the source file
f_transcurrency_a
- 转换小写金额数值为中文金额大写，负数金额时大写前加 负 字样-Convert lowercase amount value of Chinese capital, when the upper front of negative amounts of negative words
ldap
- ldap链接验证源码，已经在项目中使用。-ldap connect and check code
1001
- PAT基础题1001.害死人不偿命的(3n+1)猜想 简洁易懂 可以给大家提供一些解题思路 扩展思维-PAT basic question 1001. Killed attractiveness of (3n+1) guess concise and can give you some idea of solving the expansion of thinking
yuesefu
- 设有编号为1,2,…,n的n（n＞0）个人围成一个圈，每个人持有一个密码m。从第一个人开始报数，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。建立模型，确定存储结构。对任意n个人，密码为m，实现约瑟夫环问题。-Has numbered 1,2, ..., n of n (n> 0) individuals in a circle, each person ho
Code4
- This the code built by me. And this code is built in C++ using some loops.-This is the code built by me. And this code is built in C++ using some loops.