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  • 成績顯示三個部份abc #include<stdio.h> #include<stdlib.h> int main(void) { float gread

    成績顯示三個部份abc #include<stdio.h> #include<stdlib.h> int main(void) { float gread printf("請輸入分數\n") scanf("%f",&gread) if(gread>=80&&gread<=100) printf("成績為A\n") else if(gread>=60&&gread<=79) { printf("成績為B\n") } else if(gread>=0&&gread<60) { printf("成績為C\n") } else { printf("分數輸入錯誤\n") } system("pause") return 0 }

    标签: include stdlib float gread

    上传时间: 2014-01-15

    上传用户:waizhang

  • AR6001 WLAN Driver for SDIO installation Read Me March 26,2007 (based on k14 fw1.1) Windows CE Em

    AR6001 WLAN Driver for SDIO installation Read Me March 26,2007 (based on k14 fw1.1) Windows CE Embedded CE 6.0 driver installation. 1. Unzip the installation file onto your system (called installation directory below) 2. Create an OS design or open an existing OS design in Platform Builder 6.0. a. The OS must support the SD bus driver and have an SD Host Controller driver (add these from Catalog Items). b. Run image size should be set to allow greater than 32MB. 3. a. From the Project menu select Add Existing Subproject... b. select AR6K_DRV.pbxml c. select open This should create a subproject within your OS Design project for the AR6K_DRV driver. 4. Build the solution.

    标签: installation Windows Driver March

    上传时间: 2014-09-06

    上传用户:yuzsu

  • 编写具有如下函数原型的递归与非递归两种函数equ

    编写具有如下函数原型的递归与非递归两种函数equ,负责判断数组a与b的前n个元素值是否按下标对应完全相同,是则返回true,否则返回false。并编制主函数对它们进行调用,以验证其正确性。 bool equ(int a[], int b[], int n) 提示:递归函数中可按如下方式来分解并处理问题,先判断最后一个元素是否相同,不同则返false;相同则看n是否等于1,是则返回true,否则进行递归调用(传去实参a、b与 n-1,去判断前n-1个元素的相等性),并返回递归调用的结果(与前n-1个元素的是否相等性相同)。

    标签: equ 函数 递归 编写

    上传时间: 2014-01-18

    上传用户:love1314

  • 一个基于GTK+的单词数值计算器

    一个基于GTK+的单词数值计算器,1、 按照规则计算单词的值,如果 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 26个字母(全部用大写)的值分别为 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26,如: WINJACK这个单词的值就为:W+I+N+J+A+C+K=23+9+14+1+3+11=71% HARDWORK=H+A+R+D+W+O+R+D=8+1+18+4+23+15+18+11=98% LOVE=L+O+V+E=12+15+22+5=54% LUCK=L+U+C+K=12+21+3+11=47% ATTITUDE= A+T+T+I+T+U+D+E=1+20+20+9+20+24+4+5=100% 2、对程序的界面布局参考如下图所示,在第一个单行文本框输入一个单词,点击“计算”按钮,按照以上算法计算出该单词的值。 3、如果在最下面的单行文本框输入一个文件路径,此文件每行记录一个单词,那么经过程序计算出各个单词的值,并把结果输出到当前目录下result.txt文件中。如果文件不存在,应该提示错误。

    标签: GTK 数值 计算器

    上传时间: 2014-01-11

    上传用户:康郎

  • This document provides guidelines and describes how to easily port S60 2nd Edition C++ application

    This document provides guidelines and describes how to easily port S60 2nd Edition C++ applications to S60 3rd Edition. The document has been written based on experiences of porting regular S60 2nd Edition applications, such as the S60 Platform: POP/IMAP Example [4] that can be downloaded from Forum Nokia. Code snippets from the example are shown in Chapter 8, “Application build changes,” and in Appendix A, “Code example." In addition, Appendix B, "Commonly used functions that require capabilities," and Appendix C, "Commonly used interfaces that have been changed or removed," provide useful information on some frequently used functions and interfaces in third-party applications.

    标签: application guidelines describes document

    上传时间: 2017-01-29

    上传用户:wang5829

  • Instead of finding the longest common subsequence, let us try to determine the length of the LCS.

    Instead of finding the longest common subsequence, let us try to determine the length of the LCS. 􀂄 Then tracking back to find the LCS. 􀂄 Consider a1a2…am and b1b2…bn. 􀂄 Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. 􀂄 Case 2: am≠bn. Wehave to find the LCS of a1a2…am-1 and b1b2…bn, and a1a2…am and b b b b1b2…bn-1 Let A = a1 a2 … am and B = b1 b2 … bn 􀂄 Let Li j denote the length of the longest i,g g common subsequence of a1 a2 … ai and b1 b2 … bj. 􀂄 Li,j = Li-1,j-1 + 1 if ai=bj max{ L L } a≠b i-1,j, i,j-1 if ai≠j L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.

    标签: the subsequence determine Instead

    上传时间: 2013-12-17

    上传用户:evil

  • 复接入

    复接入,B/W双用户使用直接扩频序列 % >>>multiple access b/w 2 users using DS CDMA % >>>format is : cdmamodem(user1,user2,snr_in_dbs) % >>>user1 and user2 are vectors and they should be of equal length % >>>e.g. user1=[1 0 1 0 1 0 1] , user2=[1 1 0 0 0 1 1],snr_in_dbs=-50 % >>>or snr_in_dbs=50 just any number wud do % Waqas Mansoor % NUST , Pakistan

    标签:

    上传时间: 2014-11-22

    上传用户:zl5712176

  • sbit CS = P1^0 sbit SCK = P1^6 sbit SIN = P1^7 sbit SOUT = P1^5 sbit WP = P1^1

    sbit CS = P1^0 sbit SCK = P1^6 sbit SIN = P1^7 sbit SOUT = P1^5 sbit WP = P1^1 void XReady(void) void XSendByte(unsigned char b) unsigned char XGetByte(void) void XWriteEn(void) void XWriteDis(void) unsigned char XReadStatus(void) void XWipPoll(void) void XWriteStatus(unsigned char b) void XReadData(unsigned char *m,unsigned char x,unsigned char n) void XWriteData(unsigned char *m,unsigned char x,unsigned char n) #define XResetDog() {CS=0 CS=1 } void InitX5045(void)

    标签: sbit SOUT SCK SIN

    上传时间: 2014-01-17

    上传用户:lijinchuan

  • 【问题描述】 设计一个利用哈夫曼算法的编码和译码系统

    【问题描述】 设计一个利用哈夫曼算法的编码和译码系统,重复地显示并处理以下项目,直到选择退出为止。 【基本要求】 (1)初始化:键盘输入字符集大小n、n个字符和n个权值,建立哈夫曼树; (2)编码:利用建好的哈夫曼树生成哈夫曼编码; (3)输出编码; (4)设字符集及频度如下表: 字符:A B C D E F 频度:4 9 23 2 17 15 字符:G H I J K 频度:1 2 3 3 4

    标签: 哈夫曼算法 编码 译码

    上传时间: 2017-03-07

    上传用户:qwe1234

  • Creating barcodes in Microsoft庐 Office has never been easier. With BarCodeWiz Toolbar you can add b

    Creating barcodes in Microsoft庐 Office has never been easier. With BarCodeWiz Toolbar you can add barcodes to Microsoft庐 Office applications with a click of a button. In Microsoft庐 Word, create single barcodes, pages of labels, or mail merge documents. In Microsoft庐 Excel庐, select a range of cells and automatically convert each cell to a barcode. In Microsoft庐 Access庐, create reports with barcodes based on your data tables.

    标签: BarCodeWiz Microsoft Creating barcodes

    上传时间: 2013-12-18

    上传用户:asddsd