Sunday, September 20, 2015

C# Thread 

using System;
using System.Threading;

public class Worker
{
    // This method will be called when the thread is started.
    public void DoWork()
    {
        while (!_shouldStop)
        {
            Console.WriteLine("worker thread: working...");
        }
        Console.WriteLine("worker thread: terminating gracefully.");
    }
    public void RequestStop()
    {
        _shouldStop = true;
    }
    // Volatile is used as hint to the compiler that this data
    // member will be accessed by multiple threads.
    private volatile bool _shouldStop;
}

public class WorkerThreadExample
{
    static void Main()
    {
        // Create the thread object. This does not start the thread.
        Worker workerObject = new Worker();
        Thread workerThread = new Thread(workerObject.DoWork);

        // Start the worker thread.
        workerThread.Start();
        Console.WriteLine("main thread: Starting worker thread...");

        // Loop until worker thread activates.
        while (!workerThread.IsAlive);

        // Put the main thread to sleep for 1 millisecond to
        // allow the worker thread to do some work:
        Thread.Sleep(1);

        // Request that the worker thread stop itself:
        workerObject.RequestStop();

        // Use the Join method to block the current thread 
        // until the object's thread terminates.
        workerThread.Join();
        Console.WriteLine("main thread: Worker thread has terminated.");
    }
}
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Tuesday, September 15, 2015

Maze

문제 번호 G: [100점] 문제06

문제 번호 G: [100점] 문제06

시간 제한: 1 Sec  메모리 제한: 128 MB
제출: 764  해결 문제 수: 98
[제출][채점 상황 열람]

문제 설명

미로탈출 로봇 대회를 개최하였다. 대회에 사용되는 미로는 가로(X), 세로(Y) 100 이하의 크기이며, 로봇이 미로를 한 칸 이동하는 데는 1초가 걸린다.
 
 
로봇이 출발점에서 도착점까지 가장 빨리 이동할 경우 걸리는 시간을 구하는 프로그램을 작성하시오.
* 입출력 Template이필요한경우 C/C++ 제출은다음코드를복사하여코드를작성하시오
 
  
#include <stdio.h>
 
int maze[101][101];
int X, Y;
int sx, sy, ex, ey;
 
int main(void)
{
    int i, j, k=0;
 
    //입력받는부분
    scanf("%d %d", &X, &Y);
    scanf("%d %d %d %d", &sx, &sy, &ex, &ey);
 
    for(i=0 ; i<Y ; i++)
    {
        for(j=0 ; j<X ; j++) scanf("%1d", &maze[i][j]);
    }
 
    //여기서부터작성
 
 
 
    //출력하는부분
    printf("%d\n", k);
 
    return 0;
}
 
* 입출력 Template이필요한경우 JAVA 제출은다음코드를복사하여코드를작성하시오
  
import java.io.IOException;
import java.util.Scanner;
 
public class Main
{
      public static void main(String [] arg) throws IOException
      {
            String [] maze ;
            int X, Y;
            int sx, sy, ex, ey;
            int i, j, k=0;
          
            Scanner sc = new Scanner(System.in);
          
            //입력받는부분
            X = sc.nextInt();
            Y = sc.nextInt();
            sx = sc.nextInt();
            sy = sc.nextInt();
            ex = sc.nextInt();
            ey = sc.nextInt();
            maze = new String [Y];
            sc.nextLine();
            for(i=0 ; i<Y ; i++) maze[i] = sc.nextLine();
          
            //여기서부터작성
          
            //출력하는부분
            System.out.println(k);
 
            sc.close();
      }
}

[힌트] 
 
입력예2
8 10                                             
1 1 7 9                                
00000101
11001000
10000010
01101010
00000110
01010000
01110110
10000100
10011101
01000001
출력예2
16
 
입력예3
5 5
1 1 5 5
00000
01101
00100
01110
00000
출력예3
8
 

입력

첫줄에 미로의 크기 X, Y(1≤X, Y≤100)가 주어진다. 둘째 줄에 출발점 x, y 좌표와 도착점 x, y 좌표가 공백으로 구분하여 주어진다. 셋째 줄부터 미로의 정보가 길은 0, 벽은 1로 공백이 없이 들어온다. (주의) 좌표는 좌측상단이 가장 작은 위치이며 이 위치의 좌표는 (1,1)이다. 

출력

첫줄에 출발점에서 도착점까지 가장 빠른 시간을 출력한다.

입력 예시

8 7
1 2 7 5
11111111
00000111
10110011
10111001
10111101
10000001
11111111

출력 예시

9

도움말






[제출][채점 상황 열람]
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참외밭 Wrong Answer 

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import java.io.IOException;

import java.util.Scanner;

public class Main
{

 public static void main(String[] arg) throws IOException
 {

  int[][] a = new int[6][2];

  int K, sum = 0;

  Scanner sc = new Scanner(System.in);

  // 입력받는부분

  K = sc.nextInt();

  for (int i = 0; i < 6; i++) {

   a[i][0] = sc.nextInt();
   a[i][1] = sc.nextInt();

  }
  // 여기서부터작성

  
  int[] eastWest  = new int[3];
  int[] southNorth  = new int[3];

  int eWIdx = 0;
  int sNIdx = 0;
  for (int i = 0; i < 6; i++) {
   if(a[i][0] == 1 || a[i][0] == 2){
    eastWest[eWIdx++] = a[i][1];
   }
   if(a[i][0] == 3 || a[i][0] == 4){
    southNorth[sNIdx++] = a[i][1];
   }
  }
  
  int ewBig = 0;
  int ewBigIdx = 0;
  int ewSmallIdx = 0;
  for(int i = 0;i < eastWest.length; i++){
   if(ewBig < eastWest[i]){
    ewBig = eastWest[i];
    ewBigIdx = i;
   }
  }
  
  if(ewBigIdx == 0){
   ewSmallIdx = 1;
  }else if(ewBigIdx == 1){
   ewSmallIdx = 2;
  }else if(ewBigIdx == 2){
   ewSmallIdx = 0;
  }
  
  int ewBigValue = eastWest[ewBigIdx];
  int ewSmallValue = eastWest[ewSmallIdx];
  
//////////////////////////////////////////////////////////////  
  
  int snBig = 0;
  int snBigIdx = 0;
  int snSmallIdx = 0;
  for(int i = 0;i < southNorth.length; i++){
   if(snBig < southNorth[i]){
    snBig = southNorth[i];
    snBigIdx = i;
   }
  }
  
  if(snBigIdx == 0){
   snSmallIdx = 2;
  }else if(snBigIdx == 1){
   snSmallIdx = 0;
  }else if(snBigIdx == 2){
   snSmallIdx = 1;
  }
  
  int snBigValue = southNorth[snBigIdx];
  int snSmallValue = southNorth[snSmallIdx];
  
//  System.out.println(ewBigValue+" "+ewSmallValue);
//  System.out.println(snBigValue+" "+snSmallValue);
  
  sum = ((ewBigValue * snBigValue) - (ewSmallValue * snSmallValue)) * K; 
  // 출력하는부분
  System.out.println(sum);

  sc.close();

 }

}
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Rotating Array by a given angle(90, 180, 270, 360 degree) 

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import java.io.IOException;

import java.util.Scanner;

public class Main
{

 public static void main(String[] arg) throws IOException
 {

  int n;

  int[][] a;

  int r;

  Scanner sc = new Scanner(System.in);

  // 입력받는부분

  n = sc.nextInt();

  a = new int[n][n];

  for (int i = 0; i < n; i++)
  {
   for (int j = 0; j < n; j++)
    a[i][j] = sc.nextInt();
  }

  for (;;)
  {
   r = sc.nextInt();

   if (r == 0)
    break;

   // 여기서부터작성
   int[][] tempA = new int[n][n];
   
   int rowXDefault = 0;
   int rowYDefault = 0;
   int rowXIncrement = 0;
   int rowYIncrement = 0;
   int colXIncrement = 0;
   int colYIncrement = 0;
   
   if(r == 90){
    rowXDefault = 0;
    rowYDefault = n -1;
    rowXIncrement = -1;
    rowYIncrement = -1;
    colXIncrement = +1;
    colYIncrement = -1;
   }else if(r == 180){
    rowXDefault = n- 1;
    rowYDefault = n -1;
    rowXIncrement = -2;
    rowYIncrement = 0;
    colXIncrement = 0;
    colYIncrement = -2;
   }else if(r == 270){
    rowXDefault = n - 1;
    rowYDefault = 0;
    rowXIncrement = -1;
    rowYIncrement = +1;
    colXIncrement = -1;
    colYIncrement = -1;
   }else if(r == 360){
    //leave them
   }
   
   
   for (int i = 0; i < n; i++)
   {
    int indexX = rowXDefault;
    int indexY = rowYDefault;
    
    for (int j = 0; j < n; j++)
    {
     tempA[i + indexX][j + indexY] = a[i][j];
     
     indexX += colXIncrement;
     indexY += colYIncrement;
    
     
    }
    rowXDefault += rowXIncrement;
    rowYDefault += rowYIncrement;
   }
   
   a = tempA;
   
   // 출력하는부분
   for (int i = 0; i < n; i++)
   {
    for (int j = 0; j < n; j++)
    {
     System.out.print(a[i][j] + " ");

    }
    System.out.println();

   }
   
  }
  sc.close();
 }
 
 
}
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Labels: , ,
N*N 배열에서 각 요소에 정수가 들어 있다. 이 배열에서 합이 가장 큰 행과 열의 번호를 각각 인쇄하라.
입력예시
5
3 -5 12 3 -21
-2 11 2 -7 -11
21 -21 -35 -93 -11
9 14 39 -98 -1
-2 -2 -2 -2 -2
출력예시
2 1
 1
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10
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import java.io.IOException;

import java.util.Scanner;

public class Main

{

 public static void main(String[] arg) throws IOException

 {

  int i, j, N;


  int[][] a;

  Scanner sc = new Scanner(System.in);

  // 입력 받는 부분

  N = sc.nextInt();

  a = new int[N][N];

  for (i = 0; i < N; i++) {

   for (j = 0; j < N; j++)
    a[i][j] = sc.nextInt();

  }

  // 여기서부터 작성
  int[] row_sum = new int[N];
  int[] col_sum = new int[N];
  
  for(i = 0; i < N; i++){
   
   int row_sum_temp = 0;
   int col_sum_temp = 0;
   
   for (j = 0; j < N; j++){
    row_sum_temp += a[i][j];
    col_sum_temp += a[j][i];
   }
   row_sum[i] = row_sum_temp; 
   col_sum[i] = col_sum_temp;
  }
  
  int max_row_index = 1;
  int max_col_index = 1;
  
  int row_max = row_sum[0];
  int col_max = col_sum[0];
  
  
  for(i = 1; i < N; i++){
   if(row_max < row_sum[i]){
    max_row_index = i + 1;
    row_max = row_sum[i];
   }

  }
  
  for(i = 1; i < N; i++){
   if(col_max < col_sum[i]){
    max_col_index = i + 1;
    col_max = col_sum[i];
//    System.out.println("max_col_index : "+max_col_index);
//    System.out.println("col_max : "+col_max);
   }
  }
  // 출력 하는 부분

  System.out.print(max_row_index + " " + max_col_index);

  sc.close();

 }

}
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Computing height as stacking bowl using parenthesis, JungOl Question 2604, (http://www.jungol.co.kr/problem.php?id=2604) 

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import java.util.*;
public class Main {

 public static void main(String[] args) {
  // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        
        String str = sc.nextLine();
        
        char[] strArray = str.toCharArray();
        
        int height = 0;
        
        for(int i = 0; i < strArray.length;i++){
         
         if(i == 0){
          height += 10;
         }else if(strArray[i] == '('){
          
          if(strArray[i-1] == '('){
           height += 5;
          }else{
           height += 10;
          }
          
         }else if(strArray[i] == ')'){
          if(strArray[i-1] == ')'){
           height += 5;
          }else{
           height += 10;
          }
         }
        }
        
        System.out.println(height);
 }
 
}
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Monday, September 14, 2015

Computing the GCD(Greatest Common Divisor) and the LCM(Lowest Common Multiple) from multiple input numbers using Euclidean algorithm, JungOl Question 1002 (http://jungol.co.kr/problem.php?id=1002) 

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import java.util.*;
public class Main {

 public static void main(String[] args) {
  // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        
        int N;
        N = sc.nextInt();
        int[] input_arr = new int[N];
        
        for(int i = 0;i < N; i++){
         input_arr[i] = sc.nextInt();
        }
        
        int current_gcd = input_arr[0];
        for(int i = 0;i < N-1;i++){
         if(current_gcd > input_arr[i+1]){
          current_gcd = gcd(input_arr[i+1], current_gcd % input_arr[i+1]);
         }else{
          current_gcd = gcd(current_gcd, input_arr[i+1] % current_gcd);
         }
        }
        
        
        int current_lcm = input_arr[0];
        for(int i = 0;i < N-1;i++){
         if(current_lcm > input_arr[i+1]){
          current_lcm = lcm(current_lcm, input_arr[i+1], gcd(current_lcm, input_arr[i+1]));
         }else{
          current_lcm = lcm(current_lcm, input_arr[i+1], gcd(input_arr[i+1], current_lcm));
         }
        }
        
        System.out.println(current_gcd+" "+current_lcm);

 }
 
 static int gcd(int a, int b){
  if(b == 0){
   return a;
  }
  return gcd(b, a%b);
 }
 
 static int lcm(int a, int b, int gcd)
 {
  return a*b/gcd;
 }

}
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