C笔试题老题重发(4)

1、C+笔试题老题重发（4）24.Assig nment 2: Picture Process ingUse C+, Java, or similarIan guagesor/a ndanymiddleware such as EJB and J2EEto process a picture with a high resolutio n (3 Mega Pixelsfor example). Usesome methodologies to degrade the resolution of the picture to make it quicker for brows ing. Then d

2、ivide the degraded picture into 9sectors equally. Click any of the 9 sectors will result a detailed picture for this sector with the same resolution as that of the original picture. This assig nment is desig ned for you to dem on strate your ability to han dle pictures.25.用,|,实现一个 W0RD(2个字节)的高低位交换!i

3、nt mai n()un sig ned short a = OxABCD;un sig ned short b ;un sig ned short c ,d;b = (ac = (a >> 8) OxOOff;d = b | c;printf( n%x ,b);printf( n%x ,c);printf( n%x ,d);return 0;结果是CDAB2俩个字节是16位前八位为高位 后八位为低位 然后结合26要开辟P1,P2,P3,P4内存来做缓冲，大小自定，但这四个缓冲的 大小要一样，并且是连续的27.有一浮点型数组A,用C语言写一函数实现对浮点数组 A进行 降序排序，并输

4、出结果，要求要以数组A作为函数的入口 (建议用冒泡排序法)void BubbleSort(double arr, i nt n)int i ， j ；int exchange = 1；/ 交换标志for(i=1;iexchange=0 ；/本趟排序开始前，交换标志应为假for( j=n-1;j>=i； j )/对当前无序区Ri.n自下向上扫描if(arr j+1 > arrj) / 交换记录arrO=arr j+1 ;R0不是哨兵，仅做暂存单元arrj+1=arr j;arrj=arr0 ;exchange=1 ;/发生了交换，故将交换标志置为真if(!excha nge)/本趟排

5、序未发生交换，提前终止算法return ；en dfor(外循环)28.找错：#i nclude#i ncludeclass Baseprivate:char * n ame;public:Base(char * className)n ame = new charstrle n(className);strcpy( name, className);Base()delete n ame;char * copyName()char newn ame 256;strcpy (newn ame, n ame);return newn ame;char * getName()return n ame

6、;static void prin t(Base base)printf( n ame: %s n , base .n ame);class Subclass : public Basepublic:Subclass(char * className) : Base(className)int mai n()Base * pBase = new Subclass( test );Base:pri nt(*pBase);printf( n ame: %s n , pBase->getName();printf( new n ame: %s n , pBase->copyName();

7、return 0;29.编写一个函数，函数接收一个字符串，是由十六进制数组成的 一组字符串，函数的功能是把接到的这组字符串转换成十进制数字.并将十进制数字返回.答案：BOOL HexToDec( LPCTSTR shex,i nt idec )int i,mid;int len 二 Istrle n( shex );if( len>8 )return FALSE;mid = 0; idec = 0;for( i=0;iif( shexi>= 0 sheximid = shexi- 0else if( shexi>= a sheximid = shexi - a +10;else if( shexi>= A shexi mid = shexi - A +10;elsereturn FALSE;mididec 二idc+mid;return TRU