大数C++程序，实现加减乘除等

FPgrowth is the one of the algorithm in frequent item set mining. It is used to find the frequent item set in a database. It will give the output in tree structure format. It is more efficient than apriori algorithm because there is no candidate generation. 

一些c程序，象棋之马踏棋盘、把算术表达式转化未逆波兰表达式、保龄球计分规则算法、可进行多达50位的大整数运算（＋X）、铁路调度算法，演示了堆栈的基本用法-Some c procedures, horse riding chess board, the arithmetic expressions are not translated into Reverse Polish expression, bowling scoring rules algorithm, can be as many as 50 large integer calculations ( X), the railway scheduling algorithm, demonstrated the basic stack Usage

寻找0点或者方程求解 寻找0点或者方程求解-Looking for 0:00 or 0:00 looking equation or equations to find or equation 0:00

三元组表的方式实现两个矩阵的加法运算-Table 3 group the realization of the two Matrix addition operations

一个高速缓冲类（缓冲池）- A high speed cushion class (cushion pond)

这个卡尔可以计算加、 减法，trancpose，乘 ！

c语言实现的离散傅立叶（DFT）变换。非快速傅里叶变换，基本的傅里叶变换。-A basic Discrete Fourie Transform program, not Fast Discrete Fourie Transform

大致想法就是对于一个数abcdefg。。。比他小或等于的有abcdefg 个，其中不包括9的有（大致就是转成9进制， 如果不存在9的话转出来的大小就是小于等于abcdefg有几个十进制的数）a*9^n + b * 9 ^(n-1)....这样。但对于在高位出现过9了，那么低位的数字也就没有意义了，计算的话反而会重复，就在高位的下一位不+就好了 所以比abcdefg小的9的个数就是abcdefg-a*9^n + b * 9 ^(n-1)。。。这样多个。 一个高精度*纯洁的数字 + 高精度的加减操作就好了（因为确认了b >= a）。。 很奇怪大神们的思想都是一开始打表10^n - 9 ^ n什么的这样。。 渣渣的思路一开始就是一级一级往上乘果然跟大神们不是一个思维方式

三色球问题 若有一个口袋放有12个球，其中有3个红的，3个白的和6个黑的，从中任取8个球， 问共 这也是一个可用穷举法求解的问题。 设任取的红球个数为i, 白球个数为 j，则黑球个数为8-i-j，用count统计不同的搭配数目- Three chromospheres questions if have a pocket to put have 12 balls, including 3 red, 3 white and 6 black, no matter what takes 8 balls, asked altogether this also is may use the question which the exhaustion method solves. Supposes no matter what takes the red ball integer is i, the white ball integer is j, then the black ball integer is 8-i-j, counts differently with count matches the number