农村水电费记帐录入，统计汇总，打印输出收费单据。可预存费用，也可欠费统计。...

介绍基于ＭＡＴＬＡＢ平台的可视化编程． 重点介绍了数据和函数的可视化．-Introduction Based on MATLAB platform visual programming. Focuses on data and function visualization.

ascii->unicode->asc

基于自适应遗传算法的电液弯辊模糊控制系统 很不错的文章 对自适应遗传算法模糊控制进行的详细的分析-Based on adaptive genetic algorithm fuzzy bending electro-hydraulic control system is very good article on adaptive genetic algorithm for fuzzy control of a detailed analysis of

电脑组机程序,后台添加配件的价格与名称.前台用户选择后自动算总价-Computer Section Machine process

高尔夫手机游戏,j2me手机游戏代码,内含图片素材和源代码文件-Golf mobile phone game, j2me mobile game code, including picture material and source code files

程序很简单，玩法也很简单。能对玩家游戏时间进行统计以决出胜负-procedure is very simple to play and is also very simple. Game players can be right time for a winner and loser Statistics

game who get last 1 match lose

相当实用的flash原代码，可以提供很多信息

约瑟夫环：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。编程打印出列顺序。-Josephus : No. 1, 2, 3, ..., n n clockwise direction by individuals sitting around a circle, each holding a password (positive integers). An optional started as a positive integer reported limits on the number of m, from the first individuals to embark on the clockwise direction since the beginning of a sequence number off, the report m stop. M reported out of the list of his password as a new value m, from a clockwise direction in the next one started from a newspaper a few, like that, until the total of all listed so far. Programming print out the order.

单循环赛中选手胜负序列求解问题 名称：单循环赛中选手胜负序列求解问题 内容：有n个选手 P 1 ,P 2 ,P 3 ,… ,P n 参加了的单循环赛，每对选手之间非胜即负。现要求求出一个选手序列 P 1 ,P 2 ,P 3 ,… ,P n , 使其满足 P i 胜 P i+ 1 (i=1,… ,n-1) 。 -Single round robin tournament players winning sequence to solve the problem name: the outcome of single round-robin tournament players to solve the problem sequence elements: There are n-player P 1, P 2, P 3, ..., P n participated in a single round-robin between each pair of players Non-win or negative. Players are now required to find a sequence of P 1 " , P 2" , P 3 " , ..., P n" , to meet the P i " W P i+ 1" (i = 1, ..., n-1).