资源列表
Contest
- The 2003 ACM Mid-Central USA Programming Contest
The2003ACMMid-AtlanticUSAProgrammingContest
- The 2003 ACM Mid-Atlantic USA Programming Contest
youxianjifanzhuan
- 解决软件开发中常见的优先级反转就餐问题，很好用的程序，希望对你有开发有帮助，UCOS。-Software development to solve the priority inversion in the common dining issue, a very good program to use, want to help you develop, UCOS.
VQ
- 利用C语言实现矢量量化算法，对数据进行聚类。-Using C language vector quantization algorithm, the data clustering.
yag
- 打标机软件说明书，介绍了打标软件校正算法，及软件说明使用说明。-Marking software manual, introduced marking software correction algorithms, and software instructions instructions.
entropy
- entropy program that is a data for a test for a value of entropy
25811248DataCheck
- algorithm c4.5 a good test for a data
houzitao
- 1979年,诺贝尔奖获得者李政道教授到中国科技大学讲学,他给少年班的同学出了这样一道算术题:有5只猴子在海边发现一堆桃子,决定第二天来平分.第二天清晨,第一只猴子最早来到,它左分右分分不开,就朝海里扔了一只,恰好可以分成5份,它拿上自己的一份走了.第2,3,4,5只猴子也遇到同样的问题,采用了同样的方法,都是扔掉一只后,恰好可以分成5份.问这堆桃子至少有多少只.据说没有一个同学能当场做出答案.李教授说用常见的方法计算很繁,问题的关键在于打破常规思维. 。 -In 1979, Nobel Priz
suanshi
- 能帮你算出给的N次的X多项式之和，并输入相应的系数a0,a1,a2....an。式子形为：a0+a1*x+a2*x1+a3*x2+...+a(n+1)*xn=sum-Can help you figure out X to the N times and polynomials, and enter the corresponding coefficients a0, a1, a2 .... an. Formula of the form: a0+ a1* x+ a2* x1+ a3* x2+.
paidui
- n个人围成一个圈，每个人都按顺序编上号，从第一个人开始数起，数到3，则这个人出列，继续数到3，每数到3的这个人就出列，最后都只会剩下一个人，这个人是多少号？-n individuals in a circle, everyone is on the order code number, starting from the first few individuals, the count of 3, then this person out of the column, continue to c
Inversion
- 本程序实现求一个给定数列所有逆序个数，程序的基本思想是二分法-The program seeking to achieve a given number of series of all reverse
infomationmanagement
- 关于运筹学的介绍，概括性强，有正对性，师数学建模的好资料-about information management