創建兩個生產者進程和兩個消費者進程，生產者進程a需要生成10000個整數,每次都將自己的進程號（用getpid()函數獲得）和生成的整數放入共享內存中（共享內存大小為64Byte）。生產者b每次從26個英文字母中選一個，并將自己的進程號和選中的字母放入共享內存中，直到26個字母全部都選中。消費者進程c負責從共享內存中讀取數據生產者進程a的數據并且將這些數據寫入文件a.out。消費者進程d從共享內存讀取進程b的數據后寫入b.out中。
標簽： 進程
上傳時間： 2014-01-24
上傳用戶：王者A
本課題設計的雙機通信系統，應能完成2臺80X86PC機的串行通信。將數據從A機發送到B機，或從B機發送到A機。
標簽： 雙機通信
上傳時間： 2014-07-13
上傳用戶：stampede
This article describes how to consume Web services in Java 1.5.0 using the new JAX-WS 2.0 API (JSR 228). Developers around the world, including me, have always complained about the hard ways to work in Java to consume even a Web service as simple as adding two numbers. However, with JAX-WS 2.0 API now available in core Java in JDK 1.5.0, life is simple like never before.
標簽： describes services article consume
上傳時間： 2013-11-27
上傳用戶：kelimu
3G Mobile Open Wide Door For E-commerce The 3G mobile Internet business in 3G era will obtain rapid development, but it still cannot become the ma in 3G era. In the age of 3G speech business, but is still a subject of value-added business will have great development, the 3G mobile Internet business.
標簽： E-commerce 3G Internet business
上傳時間： 2017-03-23
上傳用戶：zjf3110
In computer vision, sets of data acquired by sampling the same scene or object at different times, or from different perspectives, will be in different coordinate systems. Image registration is the process of transforming the different sets of data into one coordinate system. Registration is necessary in order to be able to compare or integrate the data obtained from different measurements. Image registration is the process of transforming the different sets of data into one coordinate system. To be precise it involves finding transformations that relate spatial information conveyed in one image to that in another or in physical space. Image registration is performed on a series of at least two images, where one of these images is the reference image to which all the others will be registered. The other images are referred to as target images.
標簽： different computer acquired sampling
上傳時間： 2013-12-28
上傳用戶：來茴
one of video tool Skype only in windows mobile5.0 more than running, but my task is to let it run Skype in WINCE only in windows mobile5.0 more than running, but my task is to let it run in WINCE . Leave no stone unturned when I can not get windows mobile, they thought would be needed for skype all library files to my WINCE, so perhaps will be able to run skype. 1. Used depend Show skype depends on which files. In fact, not many documents, I can remember only under the following: core.dllcommctr.dllaygeshell.dllhtmlview.dll In addition to htmlview.dll, the other documents in WINCE have (if your system does not, please IMAGE Canada on the corresponding module). Therefore, I find on the Internet htmlview.dll, the results really find
標簽： running windows mobile Skype
上傳時間： 2014-01-17
上傳用戶：miaochun888
編寫一個java應用程序。用戶從鍵盤輸入一個1-9999之間的數，程序將判斷這個數是幾位數，并判斷這個數是否回文數。回文數是指將數含有的數字逆序排列后得到的數和原數相同，例如12121,4224,6778776等都是回文數。 1）程序具有判斷用戶的輸入是否為合法整數的功能。對非法輸入（例如含有字母）要進行處理。 2）要判斷輸入數的位數，并輸出相關信息。 3）要判斷是否回文數。 二、二戰期間，英國情報人員獲取德軍的一機密電報，電報的內容為： bzdz izu sxgzd vs lh ,vpzg woflsh vs vwrh vhlsddlmp glm wrw gzy vsg .gflyz gstfzu bvsg gzsd hdmlp vml lm ,hghzvy wmz hwiry mvvdgvy izd z hzd vivsg ,ltz tmlo tmlO 情報人員已經知道，這段電報的加密方式為： 1. 首先將字符串的順序顛倒。 2. 字母互換的規律為：A->Z, B-Y, C-X...X->C, Y->B, Z-A a->z, b->y, c-x...x->c, y->b, z->a. 3. 非字母字符保持不變。 請編程幫助情報人員破譯這份機密電報。給出注釋良好的源程序和程序運行后的結果。
上傳時間： 2017-06-02
上傳用戶：dengzb84
ZEUS is a family of Eulerian (grid based) Magneto-Hydrodynamic codes (MHD) for use in astrophysics, described in a series of papers by Stone and Norman (1, 2, 3). It may be used in Cartesian (XYZ), cylindrical (ZRP), and spherical (RTP) coordinates.
標簽： Magneto-Hydrodynamic astrophysics Eulerian family
上傳時間： 2013-12-19
上傳用戶：maizezhen
兩數比較大小，交換以保持a>b 如果a已經大于b則不交換 交換前后狀態都顯示
標簽： 比較
上傳時間： 2014-12-03
上傳用戶：luke5347
實驗源代碼 //Warshall.cpp #include<stdio.h> void warshall(int k,int n) { int i , j, t; int temp[20][20]; for(int a=0;a<k;a++) { printf("請輸入矩陣第%d 行元素:",a); for(int b=0;b<n;b++) { scanf ("%d",&temp[a][b]); } } for(i=0;i<k;i++){ for( j=0;j<k;j++){ if(temp[ j][i]==1) { for(t=0;t<n;t++) { temp[ j][t]=temp[i][t]||temp[ j][t]; } } } } printf("可傳遞閉包關系矩陣是:\n"); for(i=0;i<k;i++) { for( j=0;j<n;j++) { printf("%d", temp[i][ j]); } printf("\n"); } } void main() { printf("利用 Warshall 算法求二元關系的可傳遞閉包\n"); void warshall(int,int); int k , n; printf("請輸入矩陣的行數 i: "); scanf("%d",&k); 四川大學實驗報告 printf("請輸入矩陣的列數 j: "); scanf("%d",&n); warshall(k,n); }
上傳時間： 2016-06-27
上傳用戶：梁雪文以
