双端断裂事故 double end guillotine break
双端断裂事故 double end guillotine break
相关考题：
以下程序的输出结果是______。 include int add(int x, int y) { return x＋y; } doub 以下程序的输出结果是______。include＜iostream.h＞int add(int x, int y){return x+y;}double add(double x, double y){return x+y;}void main(){int a=3, b=3;double c=6.5, d=1.5;cout＜＜add(a, b)＜＜"，"＜＜add(c, d)＜＜end1;}
编写求两个双精度数之和的函数，选项中正确的是A．double add（double a,double b）{double s;s=a+b;return s; }B．double add（double a,b）{double s;s=a+b;return （s）;}C．double add（double a double b）{double s;s=a+b;returns;}D．double add（a,b）{double a,b,s;s=a+b;return （s）;}
下面程序的输出结果是【】。include include class point{double x; double y; 下面程序的输出结果是【 】。include ＜iostream.h＞include ＜math.h＞class point{double x;double y;public:point(double a, double b){x=a;y=b;}friend double distance(point a, point b) ;};double distance(point a, point b){return sqrt ((a. x-b.x) * (a. x-b.x)+ (a. y-b. y) * (a. y-b. y) );}void main(){point p1(1,2);point p2(5,2);cout＜＜distance(p1,p2)＜＜end1;}
有以下程序include using namespace std;class Complex{public:Complex(double r=0,d 有以下程序#include ＜iostream＞using namespace std;class Complex{public: Complex(double r=0,double i=0):re(r),im(i){ double real() const {return re;} double imag() const { return im; } Complex operator+(Complex c) const {return Complex(re+c.re,im+c.im);}private: double re,im;};int main(){ Complex a =Complex(1,1)+ Complex(5); cout＜＜a.real()＜＜'+'＜＜a.imag()＜＜'i'＜＜end1; retura 0;}A．6+6iB．6+1iC．1+6iD．1+1i
分析以下程序执行结果【】。 include int f (int x, int y){return x,y; } double f (d 分析以下程序执行结果【 】。include＜iostream.h＞int f (int x, int y){return x,y;}double f (double x, double y) {return x,y;}void main() {int a=4, b=6;double c=2.6, d=7.4;cout＜＜f (a, b) ＜＜","＜＜f (c, d) ＜＜end1;}
下列程序的执行结果为【 】。include class Point{public:Point(double i, double j) 下列程序的执行结果为【 】。include ＜iostream. h＞class Point{public:Point(double i, double j) { x=i; y=j;}double Area() const { return 0.0;}private:double x, y;};class Rectangle: public Point{public:Rectangle(double i, double j, double k, double 1)double Area() const {return w * h;}private:double w, h;};Rectangle: :Rectangle(double i, double j, double k. double 1): Point(i,j).{w=k, h=1}void fun(Point s){cout＜＜s. Area()＜＜end1;}void main( ){Rectangle rec(3.0, 5.2, 15.0. 25.0);fun(rec)}
写出下列程序的运行结果【】。 include void func(double x, int part1, double 写出下列程序的运行结果【 】。include＜iostream. h＞void func(double x, int part1, double part2){part1=int(x)+500:part2=(x+500-part1)*100;}void main( ){int n;double x, f;x=1001. 0103;func (x, n, f):cout＜＜"Part 1="＜＜n＜＜" , part2="＜＜f＜＜end1}
阅读以下说明和C++程序，将应填入(n)处的字句写在对应栏内。【说明】以下程序的功能是计算正方体、球体和圆柱体的表面积和体积并输出。程序由4个类组成：类cube、sphere和cylinder分别表示正方体、球体和圆柱体；抽象类 container为抽象类，提供了两个纯虚拟函数surface_area()和volum()，作为通用接口。【C++程序】include＜iostream.h＞define pi 3.1416class container｛protected:double radius;public:container(double radius) {container::radius=radius;}virtual double surface_area()=0;virtual double velum()=0;};class cube:(1){ //定义正方体类public:cube(double radius):container(radius){};double surface_area () {return 6 * radius * radius;}double volum() {return radius * radius * radius;}};class sphere:(2){ //定义球体类public:sphere(double radius): container(radius){};double surface_area() { return (3);}double volum() {return pi * radius * radius * radius * 4/3;}};class cylinder:(4){ //定义圆柱体类double height;public:cylinder(double radius,double height):container(radius){container::height=height;}double surface_are a () { return 2 * pi * radius * (height+radius); }double volum () {return (5);}};void main(){container * p;cube obj1 (5);sphere obj2(5);cylinder obj3(5,5);p=obj1;cout＜＜“正方体表面积”(＜＜p-＞surface_area()＜＜end1;cont＜＜“正方体体积”＜＜p-＞volume()＜＜end1;p=obj2;cout＜＜“球体表面积”＜＜p-＞surface_area()＜＜end1;cout＜＜“球体体积”＜＜p-＞volume()＜＜end1;p=obj3;cout＜＜“球体表面积”＜＜p-＞surface_area()＜＜end1;cout＜＜“球体体积”＜＜p-＞volume()＜＜end1;}
下列程序中，先声明一个圆类circle和一个桌子类table，另外声明一个圆桌类roundtable，它是由 circle和table两个类派生的，要求声明一个圆桌类对象，并输出圆桌的高度，面积和颜色。请填空完成程序include＜iostream.h＞include＜string.h＞class circle{double radius；public：circle(double r){radius=r；}double get_area(){return 3.416*radius*radius；}}；class table{double height；public：table(double h)＜height=h；}double get_height(){return height；}}；class roundtable：public table，public circle{char *color;public：roundtable(double h，double r，char c[])： 【 】 {color=new char[strlen(c) +1]；【 】；}；char*get_color(){return color；}}：void main(){roundtable rt(0.8，1.0，“白色”);cout＜＜"圆桌的高:"＜＜rt. get_height()＜＜end1;cout＜＜"圆桌面积:"＜＜rt.get_area()＜＜end1;cout＜＜"圆桌颜色:"＜＜n.get color()＜＜end1;}
使用VC6打开考生文件夹下的工程test18_1，此工程包含一个源程序文件(est18_1.cpp，但该程序运行有问题，请改正程序中的错误，使程序的输出结果如下：Enter x and y:-4 -5xoffset=1yoffset=1angle=45radius=1.41421源程序文件test18_1.cpp 清单如下:include ＜iostream.h＞include ＜math.h＞class point{public:void set(double ix,double iy){x=ix;y=iy;}double xoffset(){return x;}double yoffset(){return y;}double angle ( ){return (180/3.14159)*atan2(y,x);}/**************** found *******************/inline radius ( ){return sqrt(x*x+y*y);}protected:double x;double y;};void main(){point p;double x,y;cout＜＜"Enter x and y:\n";cin＞＞x＞＞y;p.set(x,y);/**************** found *******************/p.x+=5;p.y+=6;/**************** found *******************/cout＜＜"xoffset="＜＜p.xoffset()＜＜end1;cout＜＜"yoffset="＜＜p.yoffset()＜＜end1;cout＜＜"angle="＜＜p.angle()＜＜end1;cout＜＜"radius="＜＜p.radius()＜＜end1;
语句printf（"a/bre/’hi/’y///bou/n"）；的输出结果是（说明：’/b’是退格符）（）.A、a/bre/’hi/’y///bouB、a/bre/’hi/’y/bouC、re’hi’youD、abre’hi’y/bou
单选题The function of a centrifugal pump double volute casing is to （）.Areduce radial thrust to the impellerBdouble the liquid velocity through the pump when compared to a single voluteCreduce hydraulic end thrustDprovide the effect of multi-staging
单选题（） is secured by means of side-chock bolts and end-chock bolts.AThe bedplateBThe columnCThe main bearingDThe double bottom
判断题超队列是一种输出受限的双端队列，即插入限制在一端（例如end1）进行，而删除仍允许在两端进行。A对B错