完整的CRC循環冗余校驗碼程序,C語言編程
上傳時間: 2013-12-26
上傳用戶:moshushi0009
一個循環冗余校驗的程序,c語言編寫的,16位的
上傳時間: 2016-05-07
上傳用戶:TF2015
溫度華氏轉變攝氏 #include <stdio.h> #include <stdlib.h> enum x {A,B,C,D,E} int main(void) { int a=73,b=85,c=66 { if (a>=90) printf("a=A等級!!\n") else if (a>=80) printf("73分=B等級!!\n") else if (a>=70) printf("73分=C等級!!\n") else if (a>=60) printf("73分=D等級!!\n") else if (a<60) printf("73分=E等級!!\n") } { if (b>=90) printf("b=A等級!!\n") else if (b>=80) printf("85分=B等級!!\n") else if (b>=70) printf("85分=C等級!!\n") else if (b>=60) printf("85分=D等級!!\n") else if (b<60) printf("85分=E等級!!\n") } { if (c>=90) printf("c=A等級!!\n") else if (c>=80) printf("66分=B等級!!\n") else if (c>=70) printf("66分=C等級!!\n") else if (c>=60) printf("66分=D等級!!\n") else if (c<60) printf("66分=E等級!!\n") } system("pause") return 0 }
上傳時間: 2014-11-10
上傳用戶:wpwpwlxwlx
溫度華氏轉變攝氏 #include <stdio.h> #include <stdlib.h> enum x {A,B,C,D,E} int main(void) { int a=73,b=85,c=66 { if (a>=90) printf("a=A等級!!\n") else if (a>=80) printf("73分=B等級!!\n") else if (a>=70) printf("73分=C等級!!\n") else if (a>=60) printf("73分=D等級!!\n") else if (a<60) printf("73分=E等級!!\n") } { if (b>=90) printf("b=A等級!!\n") else if (b>=80) printf("85分=B等級!!\n") else if (b>=70) printf("85分=C等級!!\n") else if (b>=60) printf("85分=D等級!!\n") else if (b<60) printf("85分=E等級!!\n") } { if (c>=90) printf("c=A等級!!\n") else if (c>=80) printf("66分=B等級!!\n") else if (c>=70) printf("66分=C等級!!\n") else if (c>=60) printf("66分=D等級!!\n") else if (c<60) printf("66分=E等級!!\n") } system("pause") return 0 }
上傳時間: 2013-12-12
上傳用戶:亞亞娟娟123
跨數據庫平臺: n 支持 oracle 的OCI n 支持DB2的CLI, n 支持ODBC(通過ODBC,可支持SQL SERVER,MySQL等) Ø 跨OS平臺: n 標準C++語言,支持 unix/linux/windows Ø 使用簡單: n 只有一個頭文件 n 接口簡潔.otl_stream, otl_connect, otl_exception等就可以完成大部分工作 n 相對 ProC等嵌入式開發,代碼能相應減少 Ø 性能: n 直接訪問數據庫API接口,具有API接口的高效率,可靠性 Ø 穩定性: n 開源代碼,唯一的代碼文件otlv4.h,可以了解所有基于數據庫API的實現細節 n 從1996年開始,到今已10余年. Ø 可讀性及可維護性: n 標準C++代碼,不需要任何預處理 n 使用流的形式,輸入輸出異常簡潔 n 減少大量代碼, n 代碼結構更加簡潔 參考資料: http://otl.sourceforge.net/
上傳時間: 2017-06-14
上傳用戶:cainaifa
排序算法的性能比較 對不同類型的問題規模,測試各種介紹的排序算法的性能。統計他們的鍵值比較次數,鍵值移動次數以及運行時間并分析結果。 小型問題:N≈20 中型問題:N≈2000 大型問題:N≈200000. C 語言的32關鍵字如下: 實驗原理: 線性表的排序,快速,起泡,選擇,插入。
上傳時間: 2013-12-20
上傳用戶:lingzhichao
1、 了解系統調用pipe()的功能和實際原理 2、 編寫一段程序,使用管道實現父子進程之間的通信 a) 使用系統調用fork()創建一個子進程 b) 子進程調用函數write()向父進程發送自己的進程ID和字符串” s sending a message to parent.\n”。 c) 父進程調用函數read()通過管道讀出子進程發來的消息,將消息輸出屏幕,然后終止
上傳時間: 2013-12-16
上傳用戶:古谷仁美
This book is the most accurate and up-to-date source of information the STL currently available. ... It has an approach and appeal of its own: it explains techniques for building data structures and algorithms on top of the STL, and in this way appreciates the STL for what it is - a framework. Angelika Langer, Independent Consultant and C++ Report Columnist "A superbly authored treatment of the STL......an excellent book which belongs in any serious C++ developer s library." Jim Armstrong, President 2112 F/X, Texas. \n The C++ Standard Template Library (STL) represents a breakthrough in C++ programming techniques. With it, software developers can achieve vast improvements in the reliability of their software, and increase their own productivity.
標簽: information up-to-date available currently
上傳時間: 2015-10-31
上傳用戶:CHINA526
我用matlab寫的一個corner detector, 效果比現在流行的harris,susan,CSS等效果要好。 Algorithm is derived from: X.C. He and N.H.C. Yung, Curvature Scale Space Corner Detector with Adaptive Threshold and Dynamic Region of Support , Proceedings of the 17th International Conference on Pattern Recognition, 2:791-794, August 2004. Improved algorithm has been included in A Corner Detector based on Global and Local Curvature Properties and submitted to Optical Engineering.
標簽: detector matlab corner harris
上傳時間: 2013-12-30
上傳用戶:569342831
RSA ( Rivest Shamir Adleman )is crypthograph system that used to give a secret information and digital signature . Its security based on Integer Factorization Problem (IFP). RSA uses an asymetric key. RSA was created by Rivest, Shamir, and Adleman in 1977. Every user have a pair of key, public key and private key. Public key (e) . You may choose any number for e with these requirements, 1< e <Æ (n), where Æ (n)= (p-1) (q-1) ( p and q are first-rate), gcd (e,Æ (n))=1 (gcd= greatest common divisor). Private key (d). d=(1/e) mod(Æ (n)) Encyption (C) . C=Mª mod(n), a = e (public key), n=pq Descryption (D) . D=C° mod(n), o = d (private key
標簽: crypthograph information Adleman Rivest
上傳時間: 2017-09-01
上傳用戶:chfanjiang