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