代码搜索：malloc
找到约 10,000 项符合「malloc」的源代码
代码结果 10,000
www.eeworm.com/read/294863/8197259
c pr_loqo.c
/*
* File: pr_loqo.c
* Purpose: solves quadratic programming problem for pattern recognition
* for support vectors
*
* Author: Alex J. Smola
* Created: 10/14/97
www.eeworm.com/read/394434/8224769
c a.c
#include
#include
void main()
{
int n=0,m; //n:做++运算, m:最后一个数的值
int i=0,j=0; //数组的坐标
int fx=1; //下：1，右：2，上：3，左：4
int hang;
www.eeworm.com/read/293748/8276012
c 邮局选址.c
#include"math.h"
#include"malloc.h"
int main()
{
int n,i,j,*a,*b,*c,*d,x,y,min;
scanf("%d",&n);
a=(int *)malloc(sizeof(int)*n);
b=(int *)malloc(sizeof(int)*n);
c=(int *)malloc(sizeof
www.eeworm.com/read/393534/8277248
c pr_loqo.c
/*
* File: pr_loqo.c
* Purpose: solves quadratic programming problem for pattern recognition
* for support vectors
*
* Author: Alex J. Smola
* Created: 10/14/97
www.eeworm.com/read/293532/8287901
c pr_loqo.c
/*
* File: pr_loqo.c
* Purpose: solves quadratic programming problem for pattern recognition
* for support vectors
*
* Author: Alex J. Smola
* Created: 10/14/97
www.eeworm.com/read/293531/8287961
c pr_loqo.c
/*
* File: pr_loqo.c
* Purpose: solves quadratic programming problem for pattern recognition
* for support vectors
*
* Author: Alex J. Smola
* Created: 10/14/97
www.eeworm.com/read/292616/8345428
txt spline.txt
/* 三次样条插值计算算法 */
#include "math.h"
#include "stdio.h"
#include "stdlib.h"
/*
N：已知节点数N+1
R：欲求插值点数R+1
x，y为给定函数f(x)的节点值{x(i)} (
www.eeworm.com/read/370814/9584139
c pr_loqo.c
/*
* File: pr_loqo.c
* Purpose: solves quadratic programming problem for pattern recognition
* for support vectors
*
* Author: Alex J. Smola
* Created: 10/14/97
www.eeworm.com/read/367444/9747748
c pr_loqo.c
/*
* File: pr_loqo.c
* Purpose: solves quadratic programming problem for pattern recognition
* for support vectors
*
* Author: Alex J. Smola
* Created: 10/14/97
www.eeworm.com/read/366932/9791404
c 12cplx.c
#include "stdlib.h"
int cplx(n,m,a,b,alpha,eps,x,xx,k,s,f)
int n,m,k;
void (*s)();
double a[],b[],alpha,eps,x[],xx[],(*f)();
{ double rn(double *);
int r,g,i,j,it,kt,jt,kk;