博彦科技面试题范文

博彦科技面试题范文 博彦科技 姓名： _： 问题第一部分:(智力) 1 .a, b, c : wear cap(black or white, but not all white), a and b can see, but c is blink; a,b,c in turn answer whether he can know what color he is wearing, a said no , b said no ,c said is : Q: why c know, and what color his is? 2: 100个和尚吃100个馒头,大和尚一人吃4个,小和尚4人吃一个, 问:大,小和尚各多少人? 问题第二部分:(选择) 1: 扩展名为DLL的动态链接文件的特点是_ A. 可 _的插入到其他的源程序中使用 B. 本身是一个数据文件,可以与其他程序动态的链接使用 C. 本身可以独立运行,也可以供其他程序在运行是调用 D. 本身不能独立运行,但可以供其他程序在运行时调用 2:面向对象技术中,对象是类的实例,对象有三种成分_、属性和方法(或操作) A.标识 B、规则 C、封装 D、消息 3.程序设计语言引入“类”的概念是为了解决数据保护问题，C+语言将类的成员封装在类体之中，使之具有一定的存取规则，这些规则规定了存取类的成员的权力，其中，对于用private说明的成员，它_ A. 既能被该类的成员函数访问，又能被外界直接访问 B. 只能被该类的成员函数访问，外界不能直接访问 C.不能被该类的成员函数访问，只能外界直接访问 D.既不能被该类成员函数访问，也不能外界直接访问 4._ is a set of specifications and software that allow small programs or software ponents to work together. A. ActiveX B. XML C. HTML D. DAMS 问题第三部分:(问答) 1请简要说明如何测试下面的程序? template void OutputRoots(T a, T b , T c) T d=b * b 4*a*c; if(d0) float sqrtd=sqrt(d); count ”and” Else if (d= =0) Cout -b(2*a) endl; else cout endl -b/(2*a) endl; 2、以下是一段用C编写的函数,请指出其中存在的问题: 1 int PtintDigit(void) 2 3 char*p; 4 int i=0; 5 p=(char*) malloc (256*sizeof (char); 6 for (I=0;I=256;I+,p+)*p=I; 7 printf(Here is the numbers you need; %d ”,i) 8 问题第四部分(外语考察) 请翻译下面这段文字,中文写在后面: For Chinese applicants,the first step in applying for a visa is to make an appointment by calling 65325305 or 65322228. The problem is that the two numbers are so busy that often they cant get through . Mr Ben explained :“Since all embassies have to use 6532 telephone prefix and the number of lines is limited, the US embassy is only allotted two lines for our appointment system. Insgroupsto solve the problem, we plan to work through a contractor. Contractors dont have to use the 6532 prefix , so they can be assigned move lines and hire lots of people to answer calls . Applicants will also be able to use faxes and the Inter to make appointments, which will greatly enhance flexibility and convenience, Since the contact has not been awarded yet ,