资源列表
sor
- 求解Poisson方程的超松弛迭代求法,在Fortran 中使用-Relaxation for solving Poisson equations Iterative
zh_rfft
- 整数的傅里叶变换，参数只有两个，更易于操作-Integer Fourier transform, only two parameters, but also easy to operate
Bellman
- Bellman-Ford算法的源代码。大伙自己看吧，看不懂的allenlsy@gmail.com-code for algorithm bellman-ford. allenlsy@gmail.com
max
- c++编程 求最大值，用的是递归算法，非常简单和方便的，这是针对链表的-c++ programmer seeking maximum, using a recursive algorithm is very simple and convenient, this is for the list
Reverse-link
- 反向链接一个链表。程序完整可以直接编译运行-Reverse link of a linked list. Integrity of the process can directly compile and run
java1
- 快速排序，排序是计算机算法中很重要的一部分，本人写的不算很好但也不错。-Quick sort, sorting algorithm is a very important computer part, I write not very good but not bad.
Cows-beds
- 大一作业，奶牛床位问题。对大一的孩子来说比较难-Freshman homework, cows beds. For a child is difficult
inv
- C语言编写的矩阵求逆程序，可以实现各种阶次方阵求逆，采用的是矩阵记录方法。-Matrix inversion program written in C language, you can achieve various order matrix inverse, matrix recording method.
areaofcircle
- 一圆形游泳池如图所示，现在需在其周围建一圆形过道，并在其四周围上栅栏。 栅栏价格为35元/米，过道造价为20元/平方米。 过道宽度为3米， 游泳池半径由键盘输入。 要求编程计算并输出过道和栅栏的造价。-A circular pool as shown in figure, now in its built around to a circular corridor, and in its 4 around on the fence. The fence for 35 yuan
hermite
- programa que te permite calcular estadisticamente mediante hermite espero que os guste
yue
- 约瑟夫环解决一个历史上的问题,主要是一群人围成一个环，解决最后谁活下来的问题-Josephus solve a history of problems, mainly a group of people surrounded by a ring, to solve the last problem of who survived
frame_win
- 语音分帧程序，-Voice framing procedures,