更多"Windows provides a user-friendly gr"的相关试题:
[填空题]PowerPoint is a public-speaking application included in the Microsoft Office software package.
[填空题]以下程序的输出结果是 [9] 。
include<iostream.h>
include <string.h>
void main 0 {
char s[50];
strcpy(&s[O], "No" );
strcpy(&s[1], "123" );
strcpy (&s[2], "23456" );
cout<<s;
}
[填空题]以下程序运行后的输出结果是 【14】 。
#include <iostream>
#include <string>
using namespace std;
class Y;
class X
int x;
char *strx;
public:
X(int a,char *str)
x=a;
strx=new char[strlen(str)+1];
strcpy(strx,str);
void show(Y &ob) ;
;
class Y
private:
int y;
char *stry;
public:
Y(int b,char *str)
y=b;
stry=new char[strlen(str)+1];
strcpy(stry, str);
friend void X::show(Y &ob) ;
;
void X::show(Y &ob)
cout<<strx<<",";
cout<<ob.stry<<end1;
int main( )
X a(10,"stringX");
Y b(20,"stringY");
a. show (b) ;
return O;
[填空题]以下程序的运行结果是______。
#include<iostream.h>
#include<math.h>
template<class T>
class TAdd//定义类模板TAdd,T为类型
{
Tx,y;
public:
TAdd (Ta,Tb) {x=a,y=b;) //构造函数
Tadd( ) { retum x+y;}//成员函数
};
void main( )
{
TAdd<int>A (5,6); //定义TAdd<int>模板类的一个对象A
TAddB(2.4,5.8); //定义TAdd<double>模板类的一个对象B
cout<<"s1="<<A.add( )<<endl;
cout<<"s2="<<B.add( )<<endl;
}
[单项选择]():A graphical bar with buttons that perform some of the most common commands.
A. Title bar
B. Tool bar
C. Status bar
D. Scroll bar
[单项选择]
()is not the advantage of GERT (graphical evaluation and review techniques) as compared with PERT (project evaluation and review techniques).
A. Allowing looping
B. Allowing for dummy tasks
C. Allowing branching
D. Allowing multiple project end results
[单项选择]The criteria to determine a word’s category include all the following EXCEPT
A. inflection.
B. distribution.
C. meaning.
D. style.
[单项选择]
( )is not the advantage of GERT (Graphical evaluation and review techniques)as compared with PERT (project evaluation and review techniques).
A. Allowing looping
B. Allowing for dummy tasks
C. Allowing branching
D. Allowing multiple project end results
[简答题]在下面程序的横线处填上适当内容,使程序执行结果为:"hello,andylin"。
#include<iostream>
#include<string.h>
using namespace std;
class mystring
public:
char*pdata;
mystring(int len)
pdata=new char[len+1];
~mystring( )
delete pdata;
void show( )cout<<pdata<<endl;
;
void fun(mystring**array,int len)
mystring*old=*array;
______
memcpy(*array,old,len);
void main( )
mystring str(20);
mystring*pstr=&str;
mystring**ppstr=&pstr;
strcpy(str.pdata,"hello,andylin");
fun(ppstr,20);
______
[简答题]【说明】Point是平面坐标系上的点类,Line是从Point派生出来的直线类。
#include <iostream.h>
class Point
{public:
Point (int x, int y) ;
Point (Point &p) ;
~Point( );
void set (double x, double y) ;
void print( );
private:double X,Y;
};
Point::Point (int x, int y) //Point 构造函数
{X=x; Y=y; }
Point::Point ( (1) ) //Point 拷贝构造函数
{X=p.X; Y=p.Y;}
void Point::set (double x, double y)
{X=x; Y=y; }
void Point::print( )
{cout<<’ (’<<X<<","<<Y<<") "<<endl; }
Point::~Point( )
{cout<<"Point 的析构函数被调用! "<<endl;
class Line: public Point
{public:
Line (int x, int y, int k) ;
Line (Line &s) ;
~Line( );
void set (double x, double y, double k)
void print( );
private:double K;
};
(2) //Line 构造函数实现
{ K=k;}
(3) //Line 拷贝构造函数实现
{K=s.K;}
void Line::set (double x, double y, double k)
{ (4) ;
K=k;
}
void Line::print( )
{cout<<" 直线经过点";
(5) ;
cout<<"斜率为: k="<<K<<endl;
}
Line: :~Line( )
{cout<<
[填空题]以下程序运行时,若从键盘输入:10 20 30<回车>,输出的结果是 【6】 。
#include<stdio.h>
main( )
{ int i=0,j=0,k=0;
scanf("%d%*d%d",&i,&j,&k);
printf("%d%d%d/n",i,j,k);
}
[填空题]函数main( )的功能是:在带头结点的单链表中查找数据域中值最小的结点。请填空。
#include<stdio.h>
struct node
int data;
struct node * next;
int main(struct node * first)/*指针first为链表头指针*/
struct node * p;int m;
p=first->next;m=p->data;p=p->next;
for(; p!=NULL;p=______)
if(p->data<m)m=p->data;
return m;
[单项选择]运行以下程序后,如果从键盘上输入shose后回车,则输出结果为 _______。 # include<iostream.h> void main( ){ int v1=0,v2=0,k=0; char ch[80]; cin>>ch; while(ch[k]){ switch(ch[k]){ case ’s’; case ’h’; default: v1++1; case ’o’: v2++; } k++; } cout<<v1<< ","<<v2<<end1; }
A. 2, 0
B. 5, 0
C. 4, 5
D. 2, 5
[填空题]以下程序打开新文件f.txt,并调用字符输出函数将a数组中的字符写入其中,请填空。
#include< stdio.h>
main( )
(14) *fp;
char a[5]=’1’, ’2’, ’3’, ’4’, ’5’, i;
fp=fopen(" (15) ", "w");
for(i=0; i<5; i++)fputc(a[i], fp);
fclose(fp);
[简答题]
下面是一段C程序,其功能为计算1到所输入数值的逐步累加和,请阅读程序并完成要求。
#include<stdio.h>
int sum(int);
void main( )
int num,k;
printf("Enter a number:");
seanf("%d",&num);
for(k=1;k<=num;k++)
printf("1 to %d is %d/n",k,sum(k));
int sum(int n)
int i,result=0;
for(i=1;i<=n;i++)
result+=i;
return result;
请画出此程序各函数的控制流程图。
[单项选择]以下程序十函数scmp的功能是返回形参指针s1和s2所指字符串中较小字符串的首地址:
#include <stdio.h>
#include <string.h>
char *scrap(char *s1,char *s2)
if(strcmp(s1,s2)<0)
return(s1);
else return(s2);
main( )
int i; char string[20], str[3][20];
for(i=0;i<3;i++) gets(str[i]);
strcpy(string,scmp(str [0],str[1]));/*库函数strcpy对宁符中进行复制*/
strcpy(string,scmp(string,str[2]));
printf("%s/n",string);
若运行时依次输入:abcd、4bba和abc三个宁符市,则输㈩结果为
A. abcd
B. abba
C. abc
D. abca
[单项选择]以下程序中函数scmp功能是返回形参指针s1和s2所指字符串中较小字符串的首地址。
#includ<string. h>
char *scmp(char *s1,char *s2) if(strcmp(s1,s2)<0)
return(s1);
else return(s2);main( ) int i;char string[20],str[3][20];
for(i=0;i<3;i++) gets(str[i]);
strcpy(string,scmp(str[0],str[1])); //库函数 strcpy 对字符串进行复制
strcpy(string,scmp(stfing,str[2]));
puts(string);
若运行时依次输入: abed、abba 和abc三个字符串,则输出结果为( )。
A. abed
B. abba
C. abc
D. abca