代码搜索：Raspberry Pi
找到约 10,000 项符合「Raspberry Pi」的源代码
代码结果 10,000
www.eeworm.com/read/105048/15679427
cpp lagu.cpp
#include
#include
#define PI 3.14159265
main()
{
double x[5]={0,PI/6,PI/4,PI/3,PI/2};
double f[5]={0,0.5,sqrt(2.0)/2.0,sqrt(3.0)/2.0,1};
int i,j;
double sum=0,s,u=PI/
www.eeworm.com/read/105048/15679432
cpp newton插值法.cpp
//*************************************
//程序功能：Newton插值法 求函数的值
//
//作 者：杭永东
//
//日 期: 2002.06.10
//*************************************
#include
#include
#define pi
www.eeworm.com/read/103816/15719795
imp simpdiv.imp
//
// $Source: /home/gambit/CVS/gambit/sources/nash/simpdiv.imp,v $
// $Date: 2002/09/10 14:27:45 $
// $Revision: 1.3.2.1 $
//
// DESCRIPTION:
// Compute Nash equilibria via simplicial subdivision on
www.eeworm.com/read/103588/15728576
c p3.c
#include
int main()
{
int i,j,n,N;
double pi;
for(i=0; ;i++)
{
pi=0;
scanf("%d",&N);
if(N==0)
{
break;
}
for(n=3; ;n=n+2)
{
if((1.0/np
www.eeworm.com/read/102840/15756461
m exp2_4_.m
close all
clc
clear
%定义时间范围
t=[0:pi/20:9*pi];
hold on %允许在同一坐标系下绘制不同的图形
plot(t,sin(t),'r:*')
plot(t,cos(t))
plot(t,-cos(t),'k')
grid on %在所画出的图形坐标中添加栅格，注意用在plot之后
hold off %覆盖旧图
www.eeworm.com/read/102840/15756469
m exp2_4.m
close all
clc
clear
%定义时间范围
t=[0:pi/20:9*pi];
figure(1) %选择图像
plot(t,sin(t),'r:*')
grid on %在所画出的图形坐标中添加栅格，注意用在plot之后
grid off %删除栅格
figure(2)
plot(t,cos(t))
grid on
grid off
www.eeworm.com/read/102394/15783427
c tests.c
/* integration/tests.c
*
* Copyright (C) 1996, 1997, 1998, 1999, 2000 Brian Gough
*
* This program is free software; you can redistribute it and/or modify
* it under the terms of the GNU Genera
www.eeworm.com/read/101557/15826868
m ip_07_07.m
% MATLAB script for Illustrative Problem 7, Chapter 7.
echo on
Tb=1;
f1=1000/Tb;
f2=f1+1/Tb;
phi=pi/4;
N=5000; % number of samples
t=0:Tb/(N-1):Tb;
u1=cos(2*pi*f1*t);
u2=cos(2*pi*f2*t)
www.eeworm.com/read/101253/15839424
c 矩阵转换.c
void trans(int *p,int n)
{
int i,j,temp;
int *pi,*pj;
for(i=0;i
www.eeworm.com/read/100779/15864450
cpp dctf.cpp
#include "stdafx.h"
/*#include "dctf.h"
#include "math.h"
#define pi 3.1416926
float k(int t)
{
if(t=0)
return 1/(sqrt(2));
else
return 1;
}
void dct(int t0[][],float t1[][],int m,