代码搜索:Infinity

找到约 1,499 项符合「Infinity」的源代码

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www.eeworm.com/read/433256/8535786

h mila.h

#include #include #include const double PI=4*atan(1); const double RAD_DEG=180.0/PI; const double INIT_SEED=113; const double MIN_XY=1.0; const double MAX_XY=
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h contai2a.h

#include #include #include const double PI=4*atan(1); const double RAD2DEG=180.0/PI; const double INIT_SEED=113; const double MIN_XY=1.0; const double MAX_XY=
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h intmath.h

// IntMath.h: Header for doing fractional math with integers for speed. #ifndef IntMath_h #define IntMath_h typedef float BASETYPE; //typedef double BASETYPE; // Scaling factor used to stor
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m rctdm.m

function [ret,x0,str,ts,xts]=rctdm(t,x,u,flag); %RCTDM is the M-file description of the SIMULINK system named RCTDM. % The block-diagram can be displayed by typing: RCTDM. % % SYS=RCTDM(T,X,U,FLAG
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c ripmetric.c

/* ripmetric.c - ripmetric */ #include #include #include #ifdef RIP /*------------------------------------------------------------------------ * ripmetric -
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m trunc_d.m

% PURPOSE: demo of truncated normal draws % plots pdf for various truncated normal draws % %--------------------------------------------------- % USAGE: trunc_d %-----------------------
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c cmplx.c

/* cmplx.c * * Complex number arithmetic * This version is for C9X. * * * * SYNOPSIS: * * typedef struct { * double r; real part * double i; imaginary part *
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m example1.m

%***************************************************************% %问题描述:例1 现有4种不同的车床1,2,3,4同时加工500件相同的零件, % 各车床加工一个零件的时间分别为0.5,0.1,0.2和0.05h.问如何给4个 % 车床分配加工零件数目,使完工时间最短? %算法设计:将问题按车床编号分为4个阶段
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asv example1.asv

%***************************************************************% %问题描述:例1 现有4种不同的车床1,2,3,4同时加工500件相同的零件, % 各车床加工一个零件的时间分别为0.5,0.1,0.2和0.05h.问如何给4个 % 车床分配加工零件数目,使完工时间最短? %算法设计:将问题按车床编号分为4个阶段
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m example1.m

%***************************************************************% %问题描述:例1 现有4种不同的车床1,2,3,4同时加工500件相同的零件, % 各车床加工一个零件的时间分别为0.5,0.1,0.2和0.05h.问如何给4个 % 车床分配加工零件数目,使完工时间最短? %算法设计:将问题按车床编号分为4个阶段