代码搜索:Difference

找到约 3,389 项符合「Difference」的源代码

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h reverse_iter.h

// // Copyright 1997, 1998, 1999 University of Notre Dame. // Authors: Andrew Lumsdaine, Jeremy G. Siek, Lie-Quan Lee // // This file is part of the Matrix Template Library // // You should have recei
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cpp adjdiff1.cpp

/* The following code example is taken from the book * "The C++ Standard Library - A Tutorial and Reference" * by Nicolai M. Josuttis, Addison-Wesley, 1999 * * (C) Copyright Nicolai M. Josuttis 19
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t-riscix

# Just for these, we omit the frame pointer since it makes such a big # difference. It is then pointless adding debugging. LIBGCC2_CFLAGS=-O2 -fomit-frame-pointer $(LIBGCC2_INCLUDES) $(GCC_CFLAGS) -g
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m ex030800.m

x = rand(1,11); n = 0:10; k = 0:500; w = (pi/500)*k; X = x * (exp(-j*pi/500)).^(n'*k); % DTFT of x % signal shifted by two samples y = x; m = n+2; Y = y * (exp(-j*pi/500)).^(m'*k); % DTFT o
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m test_ks.m

function [KSmax] = test_ks(x) % % Calculates the CDF (expcdf) [y_expcdf,x_expcdf]=cdfcalc(x); % % The theoretical CDF (theocdf) is assumed to be normal % with unknown mean and sigma zSco
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hpp glc.hpp

#if ! defined( GLC_CLASS_HEADER ) #define GLC_CLASS_HEADER /* ** Author: Samuel R. Blackburn ** Internet: wfc@pobox.com ** ** Copyright, 1996-2005, Samuel R. Blackburn ** ** "You can get c
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m ex030800.m

x = rand(1,11); n = 0:10; k = 0:500; w = (pi/500)*k; X = x * (exp(-j*pi/500)).^(n'*k); % DTFT of x % signal shifted by two samples y = x; m = n+2; Y = y * (exp(-j*pi/500)).^(m'*k); % DTFT o
www.eeworm.com/read/297947/7984374

m perform_tv_hilbert_projection.m

function [u,v,px,py] = perform_tv_hilbert_projection(f,kernel,lam,options) % Aujol & Chambolle projection for solving TV-K regularization % % [u,v,px,py] = perform_tv_hilbert_projection(f,kernel,la
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t-riscix

# Just for these, we omit the frame pointer since it makes such a big # difference. It is then pointless adding debugging. LIBGCC2_CFLAGS=-O2 -fomit-frame-pointer $(LIBGCC2_INCLUDES) $(GCC_CFLAGS) -g
www.eeworm.com/read/196830/8055719

m ex030800.m

x = rand(1,11); n = 0:10; k = 0:500; w = (pi/500)*k; X = x * (exp(-j*pi/500)).^(n'*k); % DTFT of x % signal shifted by two samples y = x; m = n+2; Y = y * (exp(-j*pi/500)).^(m'*k); % DTFT o