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定步长龙格之间的区域一体化
全区间积分的定步长龙格-库塔法-regional integration between the fixed step Changlong Runge- Kutta method
- 2023-04-12 11:45:03下载
- 积分:1
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计算两个日期间的天数差
计算两个日期间的天数差-calculate the day interval between two datetime value
- 2022-04-28 19:10:44下载
- 积分:1
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LDPC 仿真代码
1. 根据度分布生产LDPC校验矩阵H;
2. 生产LDPC编码矩阵G;
3. 在高斯信道下进行仿真;
4. 可进行不同译码算法对比。
完整的C++工程文件,方便 对LDPC码进行系统的仿真分析。
- 2022-01-28 09:57:50下载
- 积分:1
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一篇关于卡尔曼滤波的英文介绍,希望对大家有用
一篇关于卡尔曼滤波的英文介绍,希望对大家有用-An article on the English introduced the Kalman filter, in the hope that useful
- 2022-07-22 13:41:35下载
- 积分:1
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最优控制 stepest 体面
案例的最优跟踪,它是使用本文根椐梯度误差最小。在使用 stepest 体面的求解算法。resurt 会找成本函数,并尽量减少控制从状态和具中脉
- 2022-03-05 19:13:04下载
- 积分:1
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一个能将阿拉伯数字转换成汉字的程序
一个能将阿拉伯数字转换成汉字的程序-one can convert Arabic characters procedures
- 2022-06-28 11:19:29下载
- 积分:1
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numerical algorithm, BreshamLine line drawing algorithms, numerical experiments,...
数值计算算法,BreshamLine画线算法,数值计算实验,比较简单-numerical algorithm, BreshamLine line drawing algorithms, numerical experiments, relatively simple
- 2022-10-07 09:15:03下载
- 积分:1
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Sicily 1009梅森素数
One of the world-wide cooperative computing tasks is the "Grand Internet Mersenne Prime Search" -- GIMPS -- striving to find ever-larger prime numbers by examining a particular category of such numbers.
A Mersenne number is defined as a number of the form (2p–1), where p is a prime number -- a number divisible only by one and itself. (A number that can be divided by numbers other than itself and one are called "composite" numbers, and each of these can be uniquely represented by the prime numbers that can be multiplied together to generate the composite number — referred to as its prime factors.)
Initially it looks as though the Mersenne numbers are all primes.
Prime
Corresponding Mersenne Number
2
4–1 = 3 -- prime
3
8–1 = 7
- 2022-03-17 05:46:40下载
- 积分:1
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c数值算法程序大全
包含几百个数值计算算法,如低通滤波,高通滤波,带通滤波,c语言源码。
- 2022-04-25 18:07:59下载
- 积分:1
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鼹鼠闯迷宫
资源描述void creatWay(int (*mg)[N],int x, int y)//在迷宫中产生一条路,使用图的深度遍历思想来实现,
{
static int dir[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};////将要走的4个方向保存在二维数组中
int zx = x*2;
int zy = y*2;
int next, turn, i;
mg[zx][zy] = 0;
if(rand()%2)
turn = 1;
else
turn = 3;
for(i=0,next=rand()%4;i
- 2022-01-26 05:46:58下载
- 积分:1