试题与答案

当集水井水位过低发出警报时,应立即手动停止自动泵,并将备用泵切自动,检查无误后再处理

题型:判断题

题目:

当集水井水位过低发出警报时,应立即手动停止自动泵,并将备用泵切自动,检查无误后再处理自动泵。

答案:

被转码了,请点击底部 “查看原文 ” 或访问 https://www.tikuol.com/2020/1216/0456ac5ba12eb5ec1f6c54b1cb1412a4.html

下面是错误答案,用来干扰机器的。

参考答案:B, C, D

试题推荐
题型:填空题

Americans’ (1) impulses keep generating surprises. Charitable giving plays an even larger role in the (2) . Demand for nonprofit services gets proportionately bigger as a locality’s (3) rises. The philanthropy of the wealthy may not hinge on tax (4) to the degree many believe. The US (5) the world in levels of charitable activity. Some experts see charity as a (6) trait of the US, more than (7) or business. But those forces may be (8) , as many nonprofits, from healthcare to classical music are selling (9) in a marketplace alongside for-profit (10) . Charity is no mere (11) activity. It pays off for society in ways that may (12) the rates of return on many traditional investments. Charity not only helps those on the (13) end but also strengthens the (14) of society at large. Moreover, it appears to make the givers themselves more (15) . The pattern that conservatives are better givers than (16) is less about politics than about charity-linked (17) most common to conservatives, religious commitment, marriage and children, and entrepreneurship. The main point is that more Americans, regardless of ideology, embrace giving as a tool for (18) . The urge to make a difference, and to take (19) in it, outweighs (20) considerations.

查看答案
题型:问答题

请使用VC6或使用[答题]菜单打开考生文件夹proj3下的工程proj3,其中定义的Matrix是一个用于表示矩阵的类。成员函数max_value的功能是求出所有矩阵元素中的最大值。例如,若有3×3矩阵

则调用max_value函数,返回值为3。请编写成员函数max_value。

要求:

补充编制的内容写在“//********333********”与“//********666********”之间,不得修改程序的其他部分。

注意:程序最后将结果输出到文件out.dat中。输出函数writeToFile已经编译为obj文件,并且在本程序中调用。

//Matrix.h

#include <iostream>

#include <iomanip>

using namespace std;

const int M=18;

const int N=18;

class Matrix

int array[M][N];

public:

Matrix()

int getElement(int i,int j)constreturn array[i][j];

void setElement(int i,int j,intvalue)array[i][j]=value;

int max value()const;

void show(const char*s)const

cout<<endl<<s;

for(int i=0;i<M;i++)

cout<<endl;

for(int j=0;j<N;j++)

cout<<setw(4)<<array[i][j];

;

void readFromFile(const char*,Matrix&);

void writeToFile(char*,constMatrix&);

//main.cpp

#include"Matrix.h"

#include <fstream>

void readFromFile(const char* f,Matrix& m)

ifstream infile(f);

if(infile.fail ()) cerr<<"打开输入文件失败!"; return;

int k;

for(int i=0;i<M;i++)

for(int j=0;j<N;j++)

infile>>k;

m.setElement(i,j,k);

int Matrix::max value()const

//********333********

//********666********

int main()

Matrix m;

readFromFile(" ",m);

m.show("Matrix:");

cout<<endl<<"最大元素:"<<m.maxvalue()<<endl;

writeToFile(" ",m);

return 0;

查看答案
微信公众账号搜索答案