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1对1直接输出

  • 有N个城市

    有N个城市,编号为0、1…N-1,每个城市之间的路径长度保存在二位数组a中,如a[i][j]表示城市i与城市j的路径长度。求某个城市到其余城市的最短路径。

    标签: 城市

    上传时间: 2014-01-05

    上传用户:徐孺

  • Euler方法解 程序

    Euler方法解 程序,使之适用于任意右端函数f,任意步长h和任意区间[to,T]。用h=1/4,1/8,1/16分别计算初值问题

    标签: Euler 程序

    上传时间: 2015-06-09

    上传用户:缥缈

  • 应用Ehlib导出显示数据的源代码

    应用Ehlib导出显示数据的源代码,一直以来delphi配合Ehlib是非常好的数据库编程排档,但是DBGRID输出到EXCEL等格式却是十分麻烦的事情。此代码应用EHLIB自身功能可直接输出各类排版格式。应用十分简单。

    标签: Ehlib 显示数据 源代码

    上传时间: 2015-06-17

    上传用户:ANRAN

  • int13技术文档

    int13技术文档,供对磁盘直接读取编程使用

    标签: int 13 文档

    上传时间: 2015-06-28

    上传用户:三人用菜

  • 暂时只支持jpeg2000支持的 cdf97 和spline53 可以这样来测试: x=imread( E:studyjpeg2000imageslena.tif ) % see the de

    暂时只支持jpeg2000支持的 cdf97 和spline53 可以这样来测试: x=imread( E:\study\jpeg2000\images\lena.tif ) % see the decomposition coefficients y=wavelift(x, 1, spl53 ) using spline 5/3 wavelet figure subplot(1,2,1) imshow(x) subplot(1,2,2) imshow(mat2gray(y)) % see the reconstruction precision yy=wavelift(x, 5) using cdf 9/7 wavelet ix=wavelift(yy,-5) inverse sum(sum((double(x)-ix).^2))

    标签: 2000 imageslena studyjpeg imread

    上传时间: 2014-01-14

    上传用户:懒龙1988

  • DIGITAL IMAGERY is pervasive in our world today. Consequently, standards for the efficient represen

    DIGITAL IMAGERY is pervasive in our world today. Consequently, standards for the efficient representation and interchange of digital images are essential. To date, some of the most successful still image compression standards have resulted from the ongoing work of the Joint Photographic Experts Group (JPEG). This group operates under the auspices of Joint Technical Committee 1, Subcommittee 29, Working Group 1 (JTC 1/SC 29/WG 1), a collaborative effort between the International Organization for Standardization (ISO) and International Telecommunication Union Standardization Sector (ITUT). Both the JPEG [1–3] and JPEG-LS [4–6] standards were born from the work of the JPEG committee. For the last few years, the JPEG committee has been working towards the establishment of a new standard known as JPEG 2000 (i.e., ISO/IEC 15444). The fruits of these labors are now coming to bear, as JPEG-2000 Part 1 (i.e., ISO/IEC 15444-1 [7]) has recently been approved as a new international standard.

    标签: Consequently efficient pervasive standards

    上传时间: 2013-12-21

    上传用户:源弋弋

  • DNA限制性图谱的绘制方案程序 %采用改进的穷举法搜索DNA片段排序

    DNA限制性图谱的绘制方案程序 %采用改进的穷举法搜索DNA片段排序,data1,data2是传进来的两组数据,由SPDP法中测得 %例如data1=[1 14 12 3 7 8 9 6 11 4 12 3 13 2 5 10] data2=[1 1 2 1 2 2 1 2 3]

    标签: DNA 绘制 方案 搜索

    上传时间: 2014-01-19

    上传用户:wxhwjf

  • QT是目前Linux上最知名的窗口环境系统

    QT是目前Linux上最知名的窗口环境系统,除了桌上型电脑之外,PDA上的Linux系统也大都是以QT当作窗口环境。 目前QT在Linux PDA上必须安装的程序包括QT/Embedded与QT/Palmtop两者,并且QT公开其原始码与发展环境。 本程序主要希望在Linux PDA上藉由RS-232通讯与手机沟通,并且能够收发简讯。 采用的发展环境为QT/Embedded 3.0.2与GSMLib 1.8,模拟程序为QT/Palmtop的模拟器。 您可以在下列网站取得相关的函式库: gsmlib-1.8.tar.gz http://www.pxh.de/fs/gsmlib/ qt-embedded-free-3.0.2.tar.gz http://www.trolltech.com/dl/ qpe-1.3.1-bin-x86.tar.gz http://www.trolltech.com/dl/ 本程序并非免费程序,版权所有,并受 GNU General Public License 保护。

    标签: Linux 窗口 环境

    上传时间: 2014-01-01

    上传用户:baiom

  • zemax源码: This DLL models an anamorphic aspheric surface. This surface is essentially an even asp

    zemax源码: This DLL models an anamorphic aspheric surface. This surface is essentially an even aspheric surface with different terms for the x and y directions. The sag is given by: Z = ((CX*x*x)+(CY*y*y)) / (1 + sqrt(1-((1+KX)*CX*CX*x*x)-((1+KY)*CY*CY*y*y))) + AR*( (1 - AP)*x*x + (1 + AP)*y*y )^2 + BR*( (1 - BP)*x*x + (1 + BP)*y*y )^3 + CR*( (1 - CP)*x*x + (1 + CP)*y*y )^4 + DR*( (1 - DP)*x*x + (1 + DP)*y*y )^5 Note the terms AR, BR, CR, and DR ... have units of length to the -3, -5, -7, and -9 power. The terms AP, BP, CP, and DP are dimensionless. The surface is rotationally symmetric only if AP = BP = CP = DP == 0 and CX = CY and KX = KY.

    标签: surface This essentially anamorphic

    上传时间: 2015-07-25

    上传用户:lht618

  • RSA公钥加密算法基于大整数因式分解困难这样的事实。 选择两个素数

    RSA公钥加密算法基于大整数因式分解困难这样的事实。 选择两个素数,p,q。(一般p,q选择很大的数) 然后计算 z=p*q f=(p-1)(q-1) 选择一个n,使gcd(n,f)=1(gcd代表greatest common divider,一般n也选择一个素数), n和z就作为公钥。 选择一个s,0<s<f,满足n*s % f=1,s就作为私钥。

    标签: RSA 加密算法 分解 整数

    上传时间: 2013-12-14

    上传用户:wxhwjf