代码搜索:note

找到约 10,000 项符合「note」的源代码

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www.eeworm.com/read/273060/10930030

plg note.plg

Build Log --------------------Configuration: Note - Win32 Debug-------------------- Command Lines Creating command line "rc.exe /l 0x804 /fo"De
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opt note.opt

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cpp note.cpp

// Note.cpp : Defines the class behaviors for the application. // #include "stdafx.h" #include "Note.h" #include "NoteDlg.h" #ifdef _DEBUG #define new DEBUG_NEW #undef THIS_FILE static cha
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dsw note.dsw

Microsoft Developer Studio Workspace File, Format Version 6.00 # WARNING: DO NOT EDIT OR DELETE THIS WORKSPACE FILE! ###############################################################################
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clw note.clw

; CLW file contains information for the MFC ClassWizard [General Info] Version=1 LastClass=CNoteDlg LastTemplate=CEdit NewFileInclude1=#include "stdafx.h" NewFileInclude2=#include "Note.h"
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h note.h

// Note.h : main header file for the NOTE application // #if !defined(AFX_NOTE_H__F2CE3D87_4EC3_11D6_B545_00E04C104C2B__INCLUDED_) #define AFX_NOTE_H__F2CE3D87_4EC3_11D6_B545_00E04C104C2B__INCLUD
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txt note.txt

{ 令ai为向量的系数 a1(p,q)+a2(p,-q)+a3(-p,q)+a4(-p,-q)+a5(q,p)+a6(q,-p)+a7(-q,p)+a8(-q,-p)=(x,y) => (a1+a2-a3-a4)p+(a5+a6-a7-a8)q=dx (1) (a5-a6+a7-a8)p+(a1-a2+a3-a4)q=dy (2) 令b1=a1-a4,b2=a2-a3,b3=a5-a8
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txt note.txt

Ural 1141 这题n小,很容易在O(n^(0.5))找到p,q. 之后解ed≡1(mod (p-1)(q-1)),在O(logn)时间里用Extended_Euclid_GCD, 最后算c^d,用倍增思想 O(logd) 这样就不time out了. 我的理解: RSA是一种public key encryption 对于每个用户有一个public key,它是公开
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txt note.txt

用容斥也可以 规定当n
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txt note.txt

{ [1 2 3 4 | / [1 2 3 -[1 2 4 3 | | | / | | [1 4 2 3 | | | / | | [3 1 2 3 | |