SRM519

	Level 1	Level 2	Level 3
Div 1	Level 1	Level 2	Level 3
Div 2	Level 1	Level 2	Level 3

Tutorials:

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Division One - Level Three:

Solution

Source Code:

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Division One - Level Two:

Solution

Source Code:

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Division One - Level One:

Solution

The same with Div2 - 900 pts.

Source Code:

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Sivision Two - Level Three:

Solution

I just followed the tutorials on official website.

Source Code:

//Wednesday, September 28, 2011 18:08 PDT
import java.util.*;
import java.util.regex.*;
import java.text.*;
import java.math.*;
import java.awt.geom.*;

public class BinaryCards
{
    public long largestNumber(long A, long B) {
        for (int i=60; i>=0; --i)   // 60 = ceil( log2(10 ^ 18) )
            if ( ((A^B) & (1L<<i) ) != 0)
                return A | ((1L<<(i+1))-1);
        return A;
    }
     
 
}
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Division Two - Level Two:

Solution

Solve the problem recursively. It's pretty natural.

Source Code:

//Wednesday, September 28, 2011 18:08 PDT
import java.util.*;
import java.util.regex.*;
import java.text.*;
import java.math.*;
import java.awt.geom.*;

public class ThreeTeleports
{
    public int shortestDistance(int xMe, int yMe, int xHome, int yHome, String[] teleports)
    {
        long dist = Math.abs(xMe - xHome) + Math.abs(yMe - yHome);
        for(int i=0; i<teleports.length; i++){
            String[] strs = teleports[i].split(" ");
            int x1 = Integer.parseInt(strs[0]);
            int y1 = Integer.parseInt(strs[1]);
            int x2 = Integer.parseInt(strs[2]);
            int y2 = Integer.parseInt(strs[3]);
            String[] teles = new String[teleports.length - 1];
            for(int j=0, k=0; j<teleports.length; j++){
                if(i!=j){
                    teles[k++] = teleports[j];
                }
            }
         
            dist = Math.min(dist, Math.abs(xMe - x1) + Math.abs(yMe - y1) + 10L + shortestDistance(x2, y2, xHome, yHome, teles));
            dist = Math.min(dist, Math.abs(xMe - x2) + Math.abs(yMe - y2) + 10L + shortestDistance(x1, y1, xHome, yHome, teles));
        }
        return (int)dist;
    }  
}
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Division Two - Level One:

Solution

Use the sum from 1 to 7, subtract every corresponding day from the sum, leave the only missing one to return.

Source Code:

//Wed Sep 28 18:19:03 PDT 2011
import java.util.*;
import java.util.regex.*;
import java.text.*;
import java.math.*;
import java.awt.geom.*;

public class WhichDay
{
    public String getDay(String[] notOnThisDay)
    {
        int sum = (1 + 7) * 7 / 2;
        for(String str : notOnThisDay){
            sum -= getNumber(str);
        }
        return getDays(sum);
    }
 
    public int getNumber(String s){
        if(s.compareTo("Sunday") == 0) return 7;
        if(s.compareTo("Saturday") == 0) return 6;
        if(s.compareTo("Friday") == 0) return 5;
        if(s.compareTo("Thursday") == 0) return 4;
        if(s.compareTo("Wednesday") == 0) return 3;
        if(s.compareTo("Tuesday") == 0) return 2;
        if(s.compareTo("Monday") == 0) return 1;
        return 0;
    }
 
    public String getDays(int n){
        if(n == 1) return "Monday";
        if(n == 2) return "Tuesday";
        if(n == 3) return "Wednesday";
        if(n == 4) return "Thursday";
        if(n == 5) return "Friday";
        if(n == 6) return "Saturday";
        if(n == 7) return "Sunday";
        return "None";
    }  
}
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