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/449694/7498010
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/449038/7519708
m chap5_6f.m
function [kx,BsJ]=pid_fm_gaf(kx,BsJ)
global rin yout timef F
a=50;b=400;
ts=0.001;
sys=tf(b,[1,a,0]);
dsys=c2d(sys,ts,'z');
[num,den]=tfdata(dsys,'v');
u_1=0;u_2=0;
y_1=0;y_2=0;
e_1=0;
www.eeworm.com/read/447444/7551061
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/446732/7569646
txt 例2.4.txt
例2.4 求1-1/2+1/3-1/4+…+1/99-1/100。
算法可以表示如下:
S1: 1=>sign
S2: 1=>sum
S3: 2=>deno
S4: (-1)×sign=>sign
S5: sign×(1/deno)=>term
S6: sum+term=>sum
S7: deno+1=>deno
S8: 若
www.eeworm.com/read/445056/7599895
m parabola.m
function root=Parabola(f,a,b,x,eps)
if(nargin==4)
eps=1.0e-4;
end
f1=subs(sym(f),findsym(sym(f)),a);
f2=subs(sym(f),findsym(sym(f)),b);
if(f1==0)
root=a;
end
if(f2==0)
root=b;
www.eeworm.com/read/444759/7607511
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/444331/7613764
m chap5_6f.m
function [kx,BsJ]=pid_fm_gaf(kx,BsJ)
global rin yout timef F
a=50;b=400;
ts=0.001;
sys=tf(b,[1,a,0]);
dsys=c2d(sys,ts,'z');
[num,den]=tfdata(dsys,'v');
u_1=0;u_2=0;
y_1=0;y_2=0;
e_1=0;
www.eeworm.com/read/443457/7632659
cpp 2115.cpp
#include
#include
#include
using namespace std;
struct Player{
char name[100];
int mm;
int ss;
int rank;
};
Player p[11];
int n ;
void ma
www.eeworm.com/read/443457/7632715
cpp 1231(dp).cpp
#include
#include
using namespace std;
int s[100000];
void main()
{
int n,max,result,sign,start,end,st,r,i;
while(cin>>n&&n){
memset(s,0,sizeof(s));
www.eeworm.com/read/443457/7632838
cpp 2098.cpp
#include
#include
int judge(int n){
int i,sign;
sign=1;
for(i=2;i