C++
1:单选(2分H) 下列程序的输出结果是( )。:X#include <iostream.h>void main(){ int a[5]={2,4,6,8,10}; int *p=a,**q=&p; cout<<*(p++)<<','<<**q;} fq?A;YZ2b_C=Ru v[i&juLP,,x:FiDjd lr=@i& z#'FjD
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11:单选(2分@) 下列程序的输出结果是( )。)$#include 〈iostream.h〉enum TEAM{Jone,Adam,Smith=10,Bob=12,Liang};void main(){ cout<<Adam<<','<<Liang<<endl;} c:ft^0=(@9tW_RA T=0EK?4,u8TnN y.4tXq9#wBT55@g JB^&_HDGB?8MJ
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17:单选(2分\) 下述程序片段的输出是( )。tzint a[3][4]={{1,2,3,4},{5,6,7,8}};int x,*p=a[0];x=(*p)*(*p+2)*(*p+4);cout<<x<<endl; pN^QMzO:jZzp m^0M-Rh3LE v[i&juLP,,x:FiDjd lr=@i&
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21:单选(2分\) #include <iostream.h>#include <math.h>class point{ public:point(double xx,double yy){x=xx;y=yy;}void getxy();friend double distance(point &a,point &b); private:double x,y;};void point::getxy(){ cout<<〃(〃<<x<<〃,〃<<y<<〃)〃<<endl;}double distance(point &a,point &b){ double dx=a.x-b.x; double dy=a.y-b.y; return sqrt(dx*dx+dy*dy);}void main(){point p1(3.0,4.0),p2(6.0,8.0);p1.getxy();p2.getxy();double d=distance(p1,p2);cout<<d<<endl;} pN^QMzO:jZzp m^0M-Rh3LE v[i&juLP,,x:FiDjd lr=@i&
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