资源列表
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
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
SPDqu
- 是双向指针，具有三个数据块，利用指针变量，指针数组等基本语法。-double list
LordOfMU
- This a class to decrypt and encrypt mu online pakcets-This is a class to decrypt and encrypt mu online pakcets
AD7715_VER_3.0
- initialize and do continious conversion mode for Analog device AD7715.its 16 bit 1ksps. code tested ok. i used it for proccess control.code stability was excellent.-initialize and do continious conversion mode for Analog device AD7715.its 16 bit 1ksp
Beibao
- 实现背包问题，在限定的总重量内，我们如何选择，才能使得物品的总价格最高-Achieve knapsack problem, within a limited total weight, how we choose to make the total price of the highest items