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Description: designed a lowpass filter to extract cos(0.1*pi) component from x(n) -designed a lowpass filter to extract cos (0.1* pi) component from x (n)
Platform: |
Size: 1024 |
Author: linhu |
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Description: extracting cos(0.3*pi*n)component by designing the bandpass filter-extracting cos (0.3* pi* n) component by designing the bandpass filter
Platform: |
Size: 1024 |
Author: linhu |
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Description: Euler函数:
m = p1^r1 * p2^r2 * …… * pn^rn ai >= 1 , 1 <= i <= n
Euler函数：
定义：phi(m) 表示小于等于m并且与m互质的正整数的个数。
phi(m) = p1^(r1-1)*(p1-1) * p2^(r2-1)*(p2-1) * …… * pn^(rn-1)*(pn-1)
= m*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pn)
= p1^(r1-1)*p2^(r2-1)* …… * pn^(rn-1)*phi(p1*p2*……*pn)
定理：若(a , m) = 1 则有 a^phi(m) = 1 (mod m) 即a^phi(m) - 1 整出m
在实际代码中可以用类似素数筛法求出
for (i = 1 i < MAXN i++)
phi[i] = i
for (i = 2 i < MAXN i++)
if (phi[i] == i)
{
for (j = i j < MAXN j += i)
{
phi[j] /= i
phi[j] *= i - 1
}
}
容斥原理：定义phi(p) 为比p小的与p互素的数的个数
设n的素因子有p1, p2, p3, … pk
包含p1, p2…的个数为n/p1, n/p2…
包含p1*p2, p2*p3…的个数为n/(p1*p2)…
phi(n) = n - sigm_[i = 1](n/pi) + sigm_[i!=j](n/(pi*pj)) - …… +- n/(p1*p2……pk)
= n*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pk)
-err
Platform: |
Size: 1024 |
Author: Campbell |
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Description: 计算全息close all clc clear
A=zeros(64)
A(15:20,20:40)=1 A(15:50,20:25)=1
A(45:50,20:40)=1 A(30:34,20:35)=1
％ ppp=exp(rand(64)*pi*2*i) A=A.*ppp
％ Author s email: zjliu2001@163.com
figure imshow(abs(A),[])
Fa=fft2(fftshift(A)) Fs=fftshift(Fa)
Am=abs(Fs) ％ amplitude
Ph=angle(Fs) ％ phase
s=11 ％ 这表示边长吗？
cgh=zeros(64*s)
th=max(max(abs(Fs)))
-CGH close all clc clearA = zeros (64) A (15:20,20:40) = 1 A (15:50,20:25) = 1 A (45:50,20:40) = 1 A (30 : 34,20:35) = 1 ppp = exp (rand (64)* pi* 2* i) A = A.* ppp Author s email: zjliu2001@163.comfigure imshow (abs (A), [] ) Fa = fft2 (fftshift (A)) Fs = fftshift (Fa) Am = abs (Fs) amplitudePh = angle (Fs) phases = 11 This means that the length was it? cgh = zeros (64* s) th = max (max (abs (Fs)))
Platform: |
Size: 3072 |
Author: ccfu |
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Description: // 入口参数：
// l: l = 0, 傅立叶变换 l = 1, 逆傅立叶变换
// il: il = 0,不计算傅立叶变换或逆变换模和幅角；il = 1,计算模和幅角
// n: 输入的点数，为偶数，一般为32，64，128，...,1024等
// k: 满足n=2^k(k>0),实质上k是n个采样数据可以分解为偶次幂和奇次幂的次数
// pr[]: l=0时，存放N点采样数据的实部
// l=1时, 存放傅立叶变换的N个实部
// pi[]: l=0时，存放N点采样数据的虚部
// l=1时, 存放傅立叶变换的N个虚部
//
// 出口参数：
// fr[]: l=0, 返回傅立叶变换的实部
// l=1, 返回逆傅立叶变换的实部
// fi[]: l=0, 返回傅立叶变换的虚部
// l=1, 返回逆傅立叶变换的虚部
// pr[]: il = 1,i = 0 时，返回傅立叶变换的模
// il = 1,i = 1 时，返回逆傅立叶变换的模
// pi[]: il = 1,i = 0 时，返回傅立叶变换的辐角
// il = 1,i = 1 时，返回逆傅立叶变换的辐角-err
Platform: |
Size: 1024 |
Author: bluefeifei |
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Description: 给定任务序列J1J2 ...Jn，假定只有一台处理机为这批作业服务。每件任务Ji给定一个时限di和对应利润Pi。只有在规定时限之内完成作业，才会得到利润Pi。通过设计合适的算法选择和安排任务子集Ji，使得J中的每个作业都能在各自的时限内完工，且使获得的利润∑Pi总和最大。
令d=max{di} 1≤i≤n，b=min{n,d}
任何最大利润的可完工子序列中的作业个数必不大于b。
-Given task sequence J1J2 ... Jn, the assumption that only one service processor for these operations. Ji each task given a time limit and the corresponding di profits Pi. Only the stipulated time-frame to complete operations, profit would have been Pi. By designing a suitable algorithm for the selection and arrangement task subset Ji, makes J each operation can be in their respective time-frame for completion, and so was the largest sum of profits ΣPi. Order d = max (di) 1 ≤ i ≤ n, b = min (n, d) the highest profits of any sequence can be completed in the number of operations will not be greater than b.
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Size: 1024 |
Author: 林培文 |
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Description: IIR数字滤波器的设计，抽样频率为2*pi*1.5*10^4,通带截止频率为2*pi*1.5*10^3-IIR digital filter design, sampling frequency of 2* pi* 1.5* 10 ^ 4, pass-band cut-off frequency of 2* pi* 1.5* 10 ^ 3
Platform: |
Size: 1024 |
Author: nanxing |
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Description: linux下多线程实现生产者消费者及用并行算法求圆周率等经典算法-linux under the multi-threaded use of consumers and producers to achieve parallel algorithm for the classical algorithm, such as pi
Platform: |
Size: 2048 |
Author: brave_gen |
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Description: 测温的PI程序，汇编语言编写的，主程序文件，很清晰易懂！-The PI measurement procedures, the compilation of languages, the main program file, it is clear and understandable!
Platform: |
Size: 3072 |
Author: 霍群海 |
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Description: 基于BP模型的神经网络模型，simulink仿真模型-BP model based on neural network model, simulink simulation model
Platform: |
Size: 20480 |
Author: 刘伟 |
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Description: PI=3.1415926535897932384626433832795
C++学习指南，非常好的-PI = 3.1415926535897932384626433832795C++ Study guide, very good
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Size: 5120 |
Author: Nelson |
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Description: 当圆柱型阵列的来波方向为0.5137（对应的pi 坐标）时阵列波束形成方向图-When a cylindrical array for DOA 0.5137 (corresponding to the coordinates of pi) when the array beamforming pattern
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Size: 1024 |
Author: 叶紫0906 |
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Description: OFDM的发射端，包括pi/4DQPSK的调制，ifft，加帧头以及组帧-The OFDM transmitter, including pi/4DQPSK modulation, ifft, plus header and Framing
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Size: 1024 |
Author: xiaoxiao |
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Description: 此程序能完成任意表达式（包括对：正弦:sinx 余弦: cosx 正切: tanx 余切: cotx 正割: secx 余割: cscx 反正弦: arcsinx 反余弦: arccosx 反正切: arctanx 反余切: arccotx 自然对数: lnx 常数 Л: pi或PI 常数 E: : e或E 变量 X: : x|X 等几乎所有三角函数以及所有算术操作符的支持）的图像显示. 用括号区分优先级.
例如像这样的表达式： y = x*2+lne*2+tanx*(1/2) +x^5 可以直接输出它的图形，并按实际准确的坐标输出.对(中学生)在分析图像时有所帮助，只要输入一个表达式，它就可以 画出其图像，并支持扩大和缩放功能。程式里自带帮助。
开发语言: C#语言, 用Stack原理实现
-err
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Size: 76800 |
Author: xuxin |
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Description: ％ Implementation of Node Reduction Algorithm
％ and calculation of equivalent pi-shaped
％ two-port network.
％ Input Parameters:
％ - Number of internal and exterior nodes: P
％ - Admittances connecting nodes- Implementation of Node Reduction Algorithm and calculation of equivalent pi-shaped two-port network. Input Parameters:- Number of internal and exterior nodes: P- Admittances connecting nodes
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Size: 1024 |
Author: allan |
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Description: #include<stdlib.h>
#include "dos.h"
#include "graphics.h"
#include "math.h"
#define PI 3.1415926
#define NULL 0
#include<stdio.h>
#include<string.h>
-# Include <stdlib.h># Include dos.h# Include graphics.h# Include math.h# Define PI 3.1415926# Define NULL 0# Include <stdio.h># include <string.h>
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Size: 22528 |
Author: 柴利博 |
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Description: 一些用API做的例子可以供大家参考下 ?恍┯肁PI做的例子可以供大家参考下 -Using API to do some examples for your reference under? Indistinct ┯肁example PI can do for your reference under
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Size: 199680 |
Author: 疯子无 |
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Description:
Platform: |
Size: 680960 |
Author: huang |
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Description: 本程序是在MATLAB平台上，利用智能优化算法遗传算法来解决01背包问题。在0 / 1背包问题中，需对容量为c 的背包进行装载。从n 个物品中选取装入背包的物品，每件物品i 的重量为wi ，价值为pi 。对于可行的背包装载，背包中物品的总重量不能超过背包的容量，最佳装载是指所装入的物品价值最高。-This procedure is in the MATLAB platform using intelligent genetic algorithm optimization algorithm to solve the 01 knapsack problem. In the 0/1 knapsack problem, there is a need for a capacity of c for the loaded backpack. From the n-items, select the items into the backpack, each item i the weight of wi, the value of pi. Feasible for loading backpack, backpack items total weight can not exceed the capacity of backpack, the best load is the value of the items loaded the highest.
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Size: 1024 |
Author: 周薇 |
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Description: music 算法仿真 三维仿真图像，可运行-music algorithm for simulation of three-dimensional simulation images, can run
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Size: 1024 |
Author: fengdong |
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