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2019年北京神舟航天軟件筆試題和面試題答案(三)

更新:2023-09-17 00:36:00 高考升學(xué)網(wǎng)

  題目:

  1、查詢(xún)身份證號(hào)碼為440401430103082的申請(qǐng)日期

  Select applay. g_applydate

  From g_cardapply applay, g_cardapplydetail detail

  Where applay. g_applyno=detail. g_applyno

  And detail. g_idcard=’ 440401430103082’

  2、將身份證號(hào)碼為440401430103082的記錄在兩個(gè)表中的申請(qǐng)狀態(tài)均改為07

  Update g_cardapply apply, g_cardapplydetail detail set applay.g_state=’07’, detail.g_state=’07’ Where applay. g_applyno=detail. g_applyno

  And detail. g_idcard=’ 440401430103082’

  3、刪除g_cardapplydetail表中所有姓李的記錄

  Delete from g_cardapplydetail where g_name like ‘李%’

  Java API運(yùn)用

  3、請(qǐng)寫(xiě)出你所知道的Java API中所提供的數(shù)據(jù)結(jié)構(gòu)模型(例如Vector),并指出各數(shù)據(jù)模型在存儲(chǔ)結(jié)構(gòu)上和使用上有什么不同。(7分)

  4、指出JDBC中三種不同類(lèi)型的Statement(Statement、PreparedStatement、CallableStatement)的用途分別是什么。(7分)

  (三) 讀程序?qū)懡Y(jié)果(10分)

  寫(xiě)出下面程序的運(yùn)行結(jié)果,下面程序有些可能根本無(wú)法通過(guò)編譯,如果無(wú)法編譯通過(guò),請(qǐng)指出錯(cuò)誤原因。

  1、寫(xiě)出下面程序的運(yùn)行結(jié)果:(3分)

  public class Test {

  public static void changeStr(String str){

  str="welcome";

  }

  public static void main(String[] args) {

  String str="1234";

  changeStr(str);

  System.out.println(str);

  }

  }

  2、寫(xiě)出下面程序的運(yùn)行結(jié)果(5分)

  class Foo{

  public static void main(String args[]){

  int x=4,j=0;

  switch(x){

  case 1:j++;

  case 2:j++;

  case 3:j++;

  case 4:j++;

  case 5:j++;

  default:j++;

  }

  System.out.println(j);

  }

  }

  (四) 代碼查錯(cuò)(10分)

  1、指出下面程序的運(yùn)行錯(cuò)誤(4分)

  public class OutClass{

  private int varInOuterClass = 0;

  public OutClass(){

  }

  public void callOutter(){

  int varInOuterMethod = 0;

  class InnerClass{

  private int varInInnerClass = 0;

  public InnerClass(){

  }

  public void print(){

  System.out.println("varInOuterClass" + varInOuterClass);

  System.out.println("varInInnerClass" + varInInnerClass);

  System.out.println("varInInnerClass" + varInOuterMethod);

  }

  }

  InnerClass inner = new InnerClass();

  inner.print();

  }

  public static void main(String[] args){

  OutClass out = new OutClass();

  out.callOutter();

  }

  }

  2、指出下面程序的運(yùn)行錯(cuò)誤(3分)

  public class Something {

  public static void main(String[] args) {

  Something s = new Something();

  System.out.println("s.doSomething() returns " + doSomething());

  }

  public String doSomething() {

  return "Do something ...";

  }

  }

  (五) 編程題(10分)

  算法設(shè)計(jì)能力測(cè)試(10分)

  1、編寫(xiě)一個(gè)類(lèi),該類(lèi)封裝了一元二次方程共有的屬性和功能,即該類(lèi)有刻畫(huà)方程系數(shù)的3個(gè)成員變量以及計(jì)算實(shí)根的方法。

  方程:求根方法為 要求:該類(lèi)的所有對(duì)象共享常數(shù)項(xiàng)。

  下面給出了您在程序中可能會(huì)使用到的功能類(lèi),及其部分接口的API文檔,在程序中可以進(jìn)行使用,

java.lang.Math
staticfloatsignum(floatf)
Returns the signum function of the argument; zero if the argument is zero, 1.0f if the argument is greater than zero, -1.0f if the argument is less than zero.
staticdoublesin(doublea)
Returns the trigonometric sine of an angle.
staticdoublesinh(doublex)
Returns the hyperbolic sine of adoublevalue.
staticdoublesqrt(doublea)
Returns the correctly rounded positive square root of adoublevalue.
staticdoubletan(doublea)
Returns the trigonometric tangent of an angle.
staticdoubletanh(doublex)
Returns the hyperbolic tangent of adoublevalue.
staticdoubletoDegrees(doubleangrad)
Converts an angle measured in radians to an approximately equivalent angle measured in degrees.
staticdoubletoRadians(doubleangdeg)
Converts an angle measured in degrees to an approximately equivalent angle measured in radians.
  參考答案

  (一)不定項(xiàng)選擇

  1, C 2, A D 3, A D 4, B 5, E 6, A C D 7, A E 8, A B 9, A B C D

  10, A B C 11, C 12, C D 13, C 14, C 15, C 16, B 17, D, 18, A B D

  19, D, 20, C D 21,B C 22, B C 23, A C E 24, A C D

  (五)編程題

  public class Equation {

  public float a;

  public float b;

  public float c;

  public double[] d;

  public Equation(float a, float b, float c) {

  this.a = a;

  this.b = b;

  this.c = c;

  this.d = new double;

  }

  boolean getRealRoot() {

  float temp = b b - 4 a c;

  if (temp < 0)

  return false;

  else {

  this.d[0] = (-b + Math.sqrt(temp)) / (2 a);

  this.d = (-b - Math.sqrt(temp)) / (2 a);

  return true;

  }

  }

  public static void main(String[] arg) {

  Equation e1 = new Equation(1, 2, 1);

  if (e1.getRealRoot()) {

  System.out.print(e1.d[0]);

  System.out.print(";");

  System.out.println(e1.d);

  } else {

  System.out.print("no real root");

  }

  }

  }

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