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Thursday, December 12, 2013

TOPCORDER SRM598 DIV2 BinPackingEasy

Problem is


Problem Statement

     Fox Ciel has some items. The weight of the i-th (0-based) item is item[i]. She wants to put all items into bins.

The capacity of each bin is 300. She can put an arbitrary number of items into a single bin, but the total weight of items in a bin must be less than or equal to 300.

You are given the int[] item. It is known that the weight of each item is between 101 and 300, inclusive. Return the minimal number of bins required to store all items.

Definition

    
Class: BinPackingEasy
Method: minBins
Parameters: int[]
Returns: int
Method signature: int minBins(int[] item)
(be sure your method is public)

Limits

    
Time limit (s): 2.000
Memory limit (MB): 64

Constraints

- item will contain between 1 and 50 elements, inclusive.
- Each element of item will be between 101 and 300, inclusive.

Examples

0)
    
{150, 150, 150, 150, 150}
Returns: 3
You have five items and each bin can hold at most two of them. You need at least three bins.
1)
    
{130, 140, 150, 160}
Returns: 2
For example, you can distribute the items in the following way:
  • Bin 1: 130, 150
  • Bin 2: 140, 160
2)
    
{101, 101, 101, 101, 101, 101, 101, 101, 101}
Returns: 5
3)
    
{101, 200, 101, 101, 101, 101, 200, 101, 200}
Returns: 6
4)
    
{123, 145, 167, 213, 245, 267, 289, 132, 154, 176, 198}
Returns: 8


and My Code is blow

public static int minBins2(int[] items){
  Arrays.sort(items);
 
  int sp = 0;
  int ep = Math.max(items.length -1, 0);
 
  int totalNum =0; 
  while(true){
   if(items[ep] > 199){
    totalNum ++;
    ep--;
   }else{
    break;
   }
  }
 
  while(ep >= sp){
   if(items[ep] + items[sp] <= 300){
    sp++;
   }
   ep--;
   totalNum++;
  }
 
  return totalNum;
 }


## Access Step ##

1. Sort Items
2. if items are bigger than 199 add totalNum
3. sum of first item and last item are bigger than 300
   last index -- and totalnum ++
.......

i think this is very easy

TOPCORDER SRM598 DIV1

The Question is


Problem Statement

     Fox Ciel received a string as a birthday present. However, the string was too long for her, so she decided to make it shorter by erasing some characters.

The erasing process will look as follows:
  1. Find the smallest i such that the i-th character and the (i+1)-th character of the string are same.
  2. If there is no such i, end the process.
  3. Remove the i-th and the (i+1)-th character of the string, and repeat from 1.


For example, if she receives "cieeilll", she will change the string as follows: "cieeilll" -> "ciilll" -> "clll" -> "cl". You are given a String s. Return the string she will get after she erases characters as described above.

Definition

    
Class: ErasingCharacters
Method: simulate
Parameters: String
Returns: String
Method signature: String simulate(String s)
(be sure your method is public)

Limits

    
Time limit (s): 2.000
Memory limit (MB): 64

Constraints

- s will contain between 1 and 50 characters, inclusive.
- Each character in s will be a lowercase letter ('a'-'z').

Examples

0)
    
"cieeilll"
Returns: "cl"
This is the example from the statement.
1)
    
"topcoder"
Returns: "topcoder"
She won't erase any characters at all.
2)
    
"abcdefghijklmnopqrstuvwxyyxwvutsrqponmlkjihgfedcba"
Returns: ""
3)
    
"bacaabaccbaaccabbcabbacabcbba"
Returns: "bacbaca"
4)
    
"eel"
Returns: "l"


My Solution is Below

public static String simulate(String str){
  int start =0;
 
  while(str.length()-1 >start){
   char firstChar = str.charAt(start);
//   System.out.println(firstChar + " : fisrt");
   char nextChar = str.charAt(start+1);
//   System.out.println(nextChar +" : next");
   if(firstChar == nextChar){
    str = str.substring(0, start) + str.substring(start+2, str.length());
//    System.out.println("complete:" + str);
    if(start !=0){
     start--;
    }
   }else{
    start++;
   }
  }
 
//  System.out.println(str);
  return str;
 }


## Access Logic ##


my access is below but i don't think this is a good sample

because i just got 144point / 250 point

so other soultions are better than my solution

i attached other solution after my solution





cieeilll 
first point is c
first+1 point is i

c and i is differenct so first point will be 2

first point is i
first+1 point is e

c and i is differenct so first point will be 3

firtst point e
first+1 point is e

e and e is the same so cut before first point "ci" and cut after firstPoint +1 to end "ill"
and concat before string and after string

this will be ciill

first point changed first-1



Otheres are Below

public static String simulate(String s){
  for(int i=0; i<s.length()-1 i++){
    if(s.charAT(i) == s.charAT(i+1)){
     s= s.substring(0, i) + s.substring(i+2, s.length);
     i=-1;
     }
  }
}



sombody use StringBuffer.delete

and sombody use Stack

many different way i learend from topcoder --V

Sunday, December 1, 2013

Make small cafe DIY(Do it yourself)~

 
My name English Name is David Jung linving in korea.
 
2013/11/29 was just one year passed from meeting my girlFriend.
 
so I wanted make her surprized and mad samll cafe for her
 
i made it for about more than 20 hours.
 
 
 



















 
complete
~
 
wow ~