Michael의 작업실

Week 6: Linear Lists - Simulated Pointer Representation

Week 5: Linear Lists - Linked Representation

1. Write code for the class ChainWithSortMethods.

ChainWithSortMethods 클래스를 만들어라

This class is a subclass of Chain and it includes the number method insertionSort, which uses insertion sort to reorder the chain elements into nondecreasing order.
이 클래스는 Chain을 상속받는다. 그리고 인자의 크기를 오름차순 정렬하는 insertionSort 메소드를 갖고있다.
Do not create new nodes or delete old ones.

노드를 새로 생성하거나 삭제하지 마라.

a) What is the worst-case time complexity of your method? How much time does your method need if the elements are already in sorted order?

최악의 경우 에 시간복잡도는 얼마인가?

이미 정리되어 있는 경우에는 시간이 얼마나 걸리는가?

b) Test the correctness of your method by compiling and then executing the code. Use your own test data (any programs with any data are allowed).

측정의 정확성을 위해 컴파일링을 하고 실행해봐라.

너만의 테스트 데이터를 만들어라

Week 5: Linear Lists - Linked Representation

1. Write code for the class ChainWithSortMethods. This class is a subclass of Chain and it includes the number method insertionSort, which uses insertion sort to reorder the chain elements into nondecreasing order. Do not create new nodes or delete old ones.

a) What is the worst-case time complexity of your method? How much time does your method need if the elements are already in sorted order?

b) Test the correctness of your method by compiling and then executing the code. Use your own test data (any programs with any data are allowed).

template <typename T>

void BubbleSort2(T arr[], const int size)

{

bool Sorted = true;

for (int i=size; i>0 && Sorted; i--)

{

Sorted = false;

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

{

if (arr[j] < arr[j-1]) {

T t=arr[j];
arr[j]=arr[j-1];
arr[j-1]=t;

Sorted = true;

}

}

}

}

약간의 트릭을 더해서 만든 정렬이다.

정렬을 한뒤에는 true를 만들어서 그 j에 대해서는 더이성 정렬을 안한다.

정렬이 거의 된 배열에서는 일반 버블보다 월등히 빠른 속도를 보이며,

임의 배열에서는 많은 속도의 향상이 있고

역순배열에서도 약간 속도 향상이 있었다.

일단 사회시간 쪽지시험보듯 빈칸채우기 문제가 있습니다.
뭐,, 간단한 것들이지만, C++을 한글로 공부했고 문제가 한글이었으면,
쉬웠을 문제지만, 문제가 영어이다보니, 한글로 공부했을때랑 어휘때문에,
해석에 애로가 꽃피더군요,,ㄷㄷ

다음은 간단한 질문(ex>string에서 두 문자열을 더하는 연산자는? 이라던가, 서로 같지 않음을 판별하는 논리연산자는 무엇인가? 같은)이 있고,

그다음 소스의 빈칸채우기,, 이거는 기존 숙제로 내주신 문제가 거의 그대로 나왔으니,, 숙제를 다시한번 점검해보세요,

마지막 하이라이트는 함수 만들기, 간단한 작업을 하는 함수(C++은 팩토리얼 함수를 구현하라더군요,,)를 직접 코딩해야됬습니다.

자바랑 C++은 담당 교수님이 같은분이니,, 같은 유형의 문제가 나올 확률이 높네요,, 자바 공부하시는 님들께 참조가 됬으면 좋겠습니다..

그리고 항상 궁금한거 있으시면 서의성 교수님이나 저에게 물어보면 친절히 답해드립니다.
H.P 010-2885-4585
nateon : dcmichael@nate.com
msn : dcmichael@live.co.kr

1. Julian dates. Suppose you would like to know how many days ago Columbus was born. It is tedious to figure this out by hand, because months have different lengths and because you have to worry about leap years. Many people, such as astronomers, who deal with dates a lot have become tired of dealing with the craziness of the calendar and instead represent days in a completely different way: the so-called Julian day number. That value is defined as the number of days that have elapsed since Jan. 1, 4713 B.C. A convenient reference point is that October 9, 1995, is Julian day 2,450,000.

Here is an algorithm to compute the Julian day number: Set jd, jm, jy to the day, month, and year. If the year is negative, add 1 to jy. (There was no year 0. Year 1 B.C. was immediately followed by year A.D. 1) If the month is larger than February, add 1 to jm. Otherwise, add 13 to jm and subtract 1 from jy. Then compute

long jul = floor(365.25 * jy) + floor(30.6001 * jm) + d + 1720995.0

We store the result in a variable of type long; simple integers may not have enough digits to hold the value. If the date was before October 15, 1582, return this value. Otherwise, perform the following correction:

int ja = 0.01 * jy;

jul = jul + 2 - ja + 0.25 * ja;

Now write a function

long julian(int year, int month, int day)

that converts a date into a Julian day number. Use that function in a program that prompts the user for a date in the past, then prints out how many days that is away from today's date.

이게 1번문제 이고

2. Write a procedure void sort2(int& a, int& b) that swaps the values of a and b if a is greater than b and otherwise leaves a and b unchanged. For example,

int u = 2;

int v = 3;

int w = 4;

int x = 1;

sort2(u, v);

/* u is still 2, v is still 3 */

sort2(w, x);

/* w is now 1, x is now 4 */
이게 2번인데,,,
1번은 무슨 juliandate라는 만년력을 구하라는 문제 같고,,
2번은 간단히 크기가 작은것을 앞으로 큰것을 뒤로 보내는 문제 같네여,, C++은 BB를 보시면 교수님의 ppt도 있고, wiley에 보시면 교재도 있으니,, 참조변수에 관한것을 좀 보시는것이 도움이 되겠네요,,

Convert from:
in
Convert to:
mm
Value:
10
10 in = 254 mm

import java.util.Scanner;

/**
   This class converts between two units.
*/
public class ConversionCalculator
{
   public static void main(String[] args)
   {
      Scanner in = new Scanner(System.in);

      System.out.println("Convert from:");
      String fromUnit = in.nextLine();

      System.out.println("Convert to:");
      String toUnit = in.nextLine();

      UnitConverter from = new UnitConverter(fromUnit);
      UnitConverter to = new UnitConverter(toUnit);

      System.out.println("Value:") ;
      double val = in.nextDouble();

      double meters = from.toMeters(val);
      double converted = to.fromMeters(meters);

      System.out.println(val+ " "+ fromUnit+ " = "+ converted
            + " "+ toUnit);
   }
}

String input = JOptionPane.showInputDialog("Enter price:");

double price = Double.parseDouble(input);

JOptionPane.showMessageDialog(null, "Price: " + price);

System.exit(0);