代码搜索：Sign
找到约 10,000 项符合「Sign」的源代码
代码结果 10,000
www.eeworm.com/read/351010/10688571
m nndrwlin.m
function h = nndrwlin(x,y,w,c)
%NNDRWLIN Neural Network Design utility function.
% NNDRWLIN(X,Y,W,C)
% X - Vector of horizontal coordinates.
% Y - Vector of vertical coordinates.
% W - Wi
www.eeworm.com/read/158963/10707630
m nndrwlin.m
function h = nndrwlin(x,y,w,c)
%NNDRWLIN Neural Network Design utility function.
% NNDRWLIN(X,Y,W,C)
% X - Vector of horizontal coordinates.
% Y - Vector of vertical coordinates.
% W - Wi
www.eeworm.com/read/421513/10733684
asm float.asm
;*******************************************************************
; FLOATING POINT
; ADDITION AND SUBTRACTION
; ROUTINE
;****
www.eeworm.com/read/350382/10746074
m 7-15.m
%例程7－15 PCM编码
%首先，A律13折线量化的Matlab程序实现如下：
function y=pcmA13(x)
%A律13折线压缩特性
x=x/max(x); % x归一化
z=sign(x); %语音信号是双极性信号，负方向对称
x=abs(x);
for
www.eeworm.com/read/276181/10758476
c 递归车厢.c
/**********递归题改为非递归题实例 车厢********/
#include
#define MAX 4
int stack[MAX],p=-1;
struct
{
int num;
int sign;
}train[MAX];
void sub()
{
int inc;
if(p==MAX-
www.eeworm.com/read/275728/10799235
m rect.m
% rect.m
% 实现一个单相不控的全桥整流器
function[sys,xo]=rect(t,x,u,flag,R1,C,R)
% 由于不是离散系统，输出参数没有包括ts
% 此外需要用户输入3个新的参数
% 交流母线电阻R1，直流滤波电容C，负载电阻R
% 根据不控全桥整流电路的原理，列写微分方程
a1=-(1/R1+1/R)/C;
a2=-1/R/C;
b1=1
www.eeworm.com/read/349311/10836564
cpp 10020.cpp
#include
using namespace std;
int a[5],tmp,t,sign[5],ans[5];
char out[] = "0+-*/";
bool used[5];
int chose[5];
void Getinfo(){
freopen("test.in","r",stdin);
freopen(
www.eeworm.com/read/419769/10838956
m p_y.m
function pc3inte=P_y(interval,len);
% 该函数对区间进行裁减即Py投影，返回裁剪后的区间信号
if sign(interval(1))==sign(interval(len))
interval=interval.*(sign(interval)==sign(interval(1)));
inte=interp1([1,len],[i
www.eeworm.com/read/274975/10842103
m nndrwlin.m
function h = nndrwlin(x,y,w,c)
%NNDRWLIN Neural Network Design utility function.
% NNDRWLIN(X,Y,W,C)
% X - Vector of horizontal coordinates.
% Y - Vector of vertical coordinates.
% W - Wi
www.eeworm.com/read/348596/10881719
txt 1679.txt
#include
#include
int dis[101];
int count[101];
bool sign[101];
int e[100][100], n;
int prim( ) {
int i, j, k, ans = 0;
memset( sign, 0, sizeof sign );
memse