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-- Copyright (C) 1991-2008 Altera Corporation -- Your use of Altera Corporation's design tools, logic functions -- and other software and tools, and its AMPP partner logic -- functions, and a

-- Copyright (C) 1991-2004 Altera Corporation -- Any megafunction design, and related netlist (encrypted or decrypted), -- support information, device programming or simulation file, and a

-- Copyright (C) 1991-2006 Altera Corporation -- Your use of Altera Corporation's design tools, logic functions -- and other software and tools, and its AMPP partner logic -- functions, and a

/* * Basing on the two theories: * 1,(x*y)%z=((x%z)*(y%z))%z * 2,(x^y)%z=((x%z)^y)%z * To solve the y=(g^x)%p problem * left shifting x until x become * 0,compute g^x by computing * k=

-- Copyright (C) 1991-2004 Altera Corporation -- Any megafunction design, and related netlist (encrypted or decrypted), -- support information, device programming or simulation file, and a

-- Copyright (C) 1991-2007 Altera Corporation -- Your use of Altera Corporation's design tools, logic functions -- and other software and tools, and its AMPP partner logic -- functions, and a

-- Copyright (C) 1991-2007 Altera Corporation -- Your use of Altera Corporation's design tools, logic functions -- and other software and tools, and its AMPP partner logic -- functions, and a

#include #include #include /* 哈夫曼编码示例程序 哈夫曼建树算法描述： 1. 将所有权值分别构造一个只有一个结点的二叉树结点，将这些结点加入集合A（哈夫曼森林） 2. 检查集合A成员的个数，如果为1，则算法结束，集合A中唯一的结点为哈夫曼树的根。 3. 从集合A中取出根结点权值最小的两棵树a, b

[a,b,c,d]=power2sys('c5mpow1') % 获得系统的状态方程 G=ss(a,b,c,d); bode(G) % 绘制系统的 Bode 图 G1=tf(G) % 获得等效传递函数模型

-- Copyright (C) 1991-2006 Altera Corporation -- Your use of Altera Corporation's design tools, logic functions -- and other software and tools, and its AMPP partner logic -- functions, and a