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Description: Status CreateSMatrix(RLSMatrix &M)
{ // 创建稀疏矩阵M
int i
Triple T
Status k
printf(\"请输入矩阵的行数,列数,非零元素数：\")
scanf(\"％d,％d,％d\",&M.mu,&M.nu,&M.tu)
M.data[0].i=0 // 为以下比较做准备
for(i=1 i<=M.tu i++)
{
do
{
Platform: |
Size: 19853 |
Author: 冰河 |
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Description: 经典c程序100例==1--10 【程序1】 题目：有1、2、3、4个数字，能组成多少个互不相同且无重复数字的三位数？都是多少？ 1.程序分析：可填在百位、十位、个位的数字都是1、2、3、4。组成所有的排列后再去 掉不满足条件的排列。 2.程序源代码： main() { int i,j,k printf(\"\\n\") for(i=1 i<5 i++) ／*以下为三重循环*/ for(j=1 j<5 j++) for (k=1 k<5 k++) { if (i!=k&&i!=j&&j!=k) /*确保i、j、k三位互不相同*/ printf(\"%d,%d,%d\\n\",i,j,k) }-classic procedures hundred cases == 1 -- 10 procedures -- a topic : 1,2,3,4 figures, the number can be formed with each other with no repeat of the triple-digit figures? How many are? 1. Program Analysis : can fill the 100, 10, 000 the number of spaces are 1,2,3,4. With all the components removed after not satisfied with the conditions. 2. Source code : main () (int i, j, k printf ( "\\ n") for (i = 1 ilt; 5 i) / * the following as the triple cycle * / for (j = 1 JLT; 5 j) for (k = 1 KLT; 5 k) (if (i! = ki! = jj! = k) / * i, j, k three disparate * / printf ( "% d,% d,% d \\ n ", i, j, k
Platform: |
Size: 304386 |
Author: 刘宋 |
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Description: 经典c程序100例==1--10 【程序1】 题目：有1、2、3、4个数字，能组成多少个互不相同且无重复数字的三位数？都是多少？ 1.程序分析：可填在百位、十位、个位的数字都是1、2、3、4。组成所有的排列后再去 掉不满足条件的排列。 2.程序源代码： main() { int i,j,k printf("\n") for(i=1 i<5 i++) ／*以下为三重循环*/ for(j=1 j<5 j++) for (k=1 k<5 k++) { if (i!=k&&i!=j&&j!=k) /*确保i、j、k三位互不相同*/ printf("%d,%d,%d\n",i,j,k) }-classic procedures hundred cases == 1-- 10 procedures-- a topic : 1,2,3,4 figures, the number can be formed with each other with no repeat of the triple-digit figures? How many are? 1. Program Analysis : can fill the 100, 10, 000 the number of spaces are 1,2,3,4. With all the components removed after not satisfied with the conditions. 2. Source code : main () (int i, j, k printf ( "\ n") for (i = 1 ilt; 5 i)/* the following as the triple cycle*/for (j = 1 JLT; 5 j) for (k = 1 KLT; 5 k) (if (i! = ki! = jj! = k)/* i, j, k three disparate*/printf ( "% d,% d,% d \ n ", i, j, k
Platform: |
Size: 304128 |
Author: 刘宋 |
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Description: Status CreateSMatrix(RLSMatrix &M)
{ // 创建稀疏矩阵M
int i
Triple T
Status k
printf("请输入矩阵的行数,列数,非零元素数：")
scanf("％d,％d,％d",&M.mu,&M.nu,&M.tu)
M.data[0].i=0 // 为以下比较做准备
for(i=1 i<=M.tu i++)
{
do
{
-Status CreateSMatrix (RLSMatrix
Platform: |
Size: 19456 |
Author: 冰河 |
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Description: 稀疏矩阵采用三元组表示。（1）求两个具有相同行列数的稀疏矩阵A和B的相加矩阵C，并输出C。（2）求出C的转置矩阵D，输出D。-The use of sparse matrix triple that. (1) for the ranks of the two with the same number of sparse matrix A and B the sum of matrix C, and output C. (2) calculated C matrix transpose of D, the output D.
Platform: |
Size: 2048 |
Author: 夏洁 |
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Description: Triple DES encryption / decryption
Platform: |
Size: 43008 |
Author: anhmafia |
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Description: 3PCF计算多线程实现
定义：
点集D、R。
定义D中的点为ai∈D，R中的点为bi∈R。
距离：r1、r2、r3、err
求：
满足以下条件的三元组（空间中三角形）的数目
<ai, bm, bn>，|ai-bm|=r1±err且|ai-bn|=r2±err且|bm-bn|=r3±err
原始解法：
对于D中每一点ai，在R中找到与之距离为r1的点集R’，找到与之距离为r2的点集R’’。在点集R’与R’’中，查找两点间距离为r3的点组数目。累加。 -3PCF Multithread computing definition: point sets D, R. The point of the definition of D ai ∈ D, R in the point of bi ∈ R. Distance: r1, r2, r3, err requirements: the triple meet the following conditions (space triangle) the number of <ai, bm, bn> , | Ai-bm | = r1 ± err and | ai-bn | = r2 ± err and | bm-bn | = r3 ± err original solution: For D, each point ai, found in R with the distance r1 point set R ' , find the point with a distance r2 set R' ' . In the point set R ' and R' ' , find the distance between two points in the point group for the number of r3. Accumulation.
Platform: |
Size: 1024 |
Author: Junki Lee |
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Description: 一种纠3错BCH译码器的FPGA设计文章基于一种较新颖的纠3错BcH码逐步译码算法和结构原型，提出了BCH
译码器的完整实用化结构，采用FPeA设计并实现了纠3错BCH(31，16)译码器。该译码
方案的特点是主体结构通用、资源占用少、运行速度高，非常适合于需要对传输帧的帧头实
施特殊保护的数据传输应用场合。
主题词壁垒旦堡璺塑三堡￡里堡垒
O 引-Based on a noVel step．by—step decoding algorithm and its stnIcture prototype fh triple-锄rcorrecting
BCH codes，a complete aIld practical stlllcture for BCH decoder is pre∞nted and锄n’GA-based
BCH(3l，16)decoder designed．The decoding 8cheme is ch舢氇cteri既d by unive玛al core structure，哪all
device utilization cost蛐d higll speed，and thus are very suitabk f．or tIIe data tmn8mission印plic砒io腿
where the fj锄舱header needs a special protection f如m tlle comlption by IIoises．
Subject Te哪BCH decoding ％ple—error-correcting FPGA
万方
Platform: |
Size: 185344 |
Author: 谢先念 |
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Description: c语言经典算法
【程序1】
题目：有1、2、3、4个数字，能组成多少个互不相同且无重复数字的三位数？都是多少？
1.程序分析：可填在百位、十位、个位的数字都是1、2、3、4。组成所有的排列后再去
掉不满足条件的排列。
2.程序源代码：
main()
{
int i,j,k
printf("\n")
for(i=1 i<5 i++) ／*以下为三重循环*/
for(j=1 j<5 j++)
for (k=1 k<5 k++)
{
if (i!=k&&i!=j&&j!=k) /*确保i、j、k三位互不相同*/
printf(" d, d, d\n",i,j,k)
}
}
-classical algorithm c language program 1】 【Title: There are numbers 1,2,3,4, how many can be composed of distinct three-digit numbers with no repetition? Is how much? 1. Program Analysis: fill in the hundred, ten, a bit of the numbers are 1,2,3,4. Removed after the composition does not satisfy all the conditions of the arrangement of the arrangement. 2. Program source code: main () {int i, j, k printf (" \ n" ) for (i = 1 i < 5 i++) /* The following is the triple loop*/for (j = 1 j < 5 j++) for (k = 1 k < 5 k++) {if (i! = k & & i! = j & & j! = k) /* ensure i, j, k three distinct*/printf ( " d, d, d \ n" , i, j, k) }}
Platform: |
Size: 70656 |
Author: zhang |
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Description: C#实现的三重滤网交易策略，可在Wealth-lab及其他C#策略平台下应用。-Elder Triple Screen
An implementation of the "Triple Screen" technique published by Dr. Alexander Elder in 1985. The three screens that this particular implementation relies on are:
13 Week Exponential Moving Average to determine the long term trend
Stochastic D to determine short term overbought/oversold level, and K crossing D to initiate the trade setup
A buy/short stop to enter the market above the current trading range
Platform: |
Size: 2048 |
Author: kong xiangwen |
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Description: 实现手动生成一个n×m矩阵的迷宫，寻找一条从入口点到出口点的通路。迷宫有很多的分岔口，搜索整个迷宫，实际上这是对栈和回溯算法的综合操作。
要实现的主要功能：
（1）创建迷宫，将输入的迷宫数据存储到二维数组中。
（2）显示迷宫，将创建的迷宫以方阵的形式显示出来。
（3）查找路径，分别用非递归算法和递归算法查找迷宫路径。
（4）输出路径，用三元组（i,j,d）的形式输出和方阵的形式输出。
（5）.恢复迷宫，把探索过后的迷宫数据恢复。
-Manually generate a nm matrix maze to find a path from the entry point to the exit point. Maze with a lot of the fork, search the entire maze, which is actually the integrated operation of the stack and backtracking algorithm. To achieve the main functions: (1) create a maze, the input the maze data storage to a two-dimensional array. (2) shows the maze, will create the maze to the square in the form displayed. (3) Find the path, respectively, with a non-recursive algorithm and the recursive algorithm to find the maze path. (4) output path, with a triple (i, j, d) in the form of output and the form of a square of the output. (5). The recovery maze, Discover after the maze data recovery.
Platform: |
Size: 81920 |
Author: 罗裕君 |
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Description: 在许多的单片机应用系统中，都需要A/D转换器，将模拟量转换成数字量。在STM8单片机中，提供的是10位的A/D，通道数随芯片不同而不同，少的有4个通道，多的则有16个通道。
下面的实验程序首先对A/D输入进行采样，然后将采样结果的高8位（丢弃最低的2位），作为延时参数去调用延时子程序，然后再去驱动LED控制信号。因此不同的采样值，决定了LED的闪烁频率。当旋转ST三合一开发板上的电位器时，可以看到LED的闪烁频率发生变化。
-In many microcontroller applications require the A/D converter to convert analog into digital. STM8 microcontroller, providing a 10-bit A/D channel number with the chip, little has four channels, more than 16 channels. The following experimental procedure first sampling A/D input, then the sampling results 8 (drop the lowest 2), as the delay parameter to call the delay subroutine, and then go drive LED control signal. Different sampling values, the LED flashing frequency. When the potentiometer rotation ST triple development board, you can see the LED flashing frequency change.
Platform: |
Size: 4096 |
Author: tony |
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Description: 1.采用链式结构实现任意多项式的存储，求两个多项式的和。
2.假设自上而下按层次，自左至右输入每个结点的一个三元组(D, P, L/R)。其中D为本结点的元素，P为其父结点，L指示D为P 的左孩子，R指示D为P的右孩子。试写一个建立二叉树在内存的双链表示算法，并实现先根、中根、后根以及层序遍历算法。
3.采用邻接矩阵实现有向网的存储，建立有向网，并实现单源最短路径算法。-1 using an arbitrary polynomial chain structure of storage, and the sum of two polynomials. 2 Assuming a hierarchical top-down, from left to right input of each node a triple (D, P, L/R). Where D is an element node, P is the parent node, L indicates D is the left child of P, R indicates D is P' s right child. Try to write a binary tree in memory to establish a double-stranded representation algorithm and achieve first root, the root, root, and layer after traversal algorithms. 3 using adjacency matrix used to achieve network storage, to establish a network and to achieve single-source shortest path algorithm.
Platform: |
Size: 7168 |
Author: 白杨 |
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