代码搜索：sparse
找到约 3,324 项符合「sparse」的源代码
代码结果 3,324
www.eeworm.com/read/193277/8243410
m sparse.m
function Y=sparse(varargin)
%SPARSE (overloaded)
% Author Johan L鰂berg
% $Id: sparse.m,v 1.6 2006/07/26 20:17:58 joloef Exp $
if nargin < 3
error('At-least 3 arguments needed');
end
www.eeworm.com/read/192744/8287201
txt sparse.txt
Sub SPARSE(B(), N, X(), RSQ)
NMAX = 500
EPS = 0.000001
Dim G(500), H(500), XI(500), XJ(500)
EPS2 = N * EPS ^ 2
IRST = 0
1 IRST = IRST + 1
Call ASUB(X(), XI())
RP
www.eeworm.com/read/171045/9774325
h sparse.h
#include "main.h"
#define Mod2sparse_block 10 // Number of entries to block together for memory allocation
/*Structure representing a non-zero entry,
or the header for a row or column*/
type
www.eeworm.com/read/171045/9774365
cpp sparse.cpp
#include "main.h"
#include "sparse.h"
/* the details of the functions below can be refered to
Software for Low Density Parity Check Codes*/
static mod2entry *alloc_entry( mod2sparse *m)
{
mod
www.eeworm.com/read/270032/11050717
c sparse.c
/*
* Author: Jim Arvo
*
* Corrected 11/05/91 to remove a redundant form. The "C1" form as it
* appeared in GemsII was unnecessary.
*/
#include "GraphicsGems.h"
#define P1 (1
www.eeworm.com/read/148172/12487943
zip sparse.zip
www.eeworm.com/read/247775/12619190
cpp sparse.cpp
void sparse(double b[], int n, double x[], double rsq)
{
const int nmax = 500;
double eps2,eps = 0.000001;
double g[500], h[500], xi[500], xj[500];
int j;
eps2 = n * pow(eps, 2)
www.eeworm.com/read/135438/13930909
txt sparse.txt
procedure SPARSE(B:array of real; N:integer;var X:array of real; RSQ:real);
label 1;
const
NMAX = 500; EPS = 0.000001;
var
G,H,XI,XJ:array[0..500] of real;
J,ITER,IRST:integer;
www.eeworm.com/read/132413/14090084
c sparse.c
#define EPS 1.0e-6
#define FREERETURN {free_vector(xj,1,n);free_vector(xi,1,n);\
free_vector(h,1,n);free_vector(g,1,n);return;}
void sparse(b,n,x,rsq)
float b[],x[],*rsq;
int n;
{
int j,ite
www.eeworm.com/read/234916/14094004
cpp sparse.cpp
void sparse(double b[], int n, double x[], double rsq)
{
const int nmax = 500;
double eps2,eps = 0.000001;
double g[500], h[500], xi[500], xj[500];
int j;
eps2 = n * pow(eps, 2)