BlogJava-Fantasy's World-随笔分类-新闻http://www.blogjava.net/FinalFantasy/category/3676.html世界的小世界，我的大世界^_^zh-cnSat, 03 Mar 2007 09:06:44 GMTSat, 03 Mar 2007 09:06:44 GMT60Google编程挑战赛250分题目及答案http://www.blogjava.net/FinalFantasy/archive/2005/12/14/23893.htmlFinalFantasyFinalFantasyWed, 14 Dec 2005 08:32:00 GMThttp://www.blogjava.net/FinalFantasy/archive/2005/12/14/23893.htmlhttp://www.blogjava.net/FinalFantasy/comments/23893.htmlhttp://www.blogjava.net/FinalFantasy/archive/2005/12/14/23893.html#Feedback5http://www.blogjava.net/FinalFantasy/comments/commentRss/23893.htmlhttp://www.blogjava.net/FinalFantasy/services/trackbacks/23893.html

Problem Statement

 

    

When a stone is thrown across water, sometimes it will land on the water and bounce rather than falling in right away. Suppose that a stone is thrown a distance of n. On each successive bounce it will travel half the distance as the previous bounce (rounded down to the nearest integer). When it can not travel any further, it falls into the water. If, at any point, the stone lands on an obstruction rather than water, it will not bounce, but will simply deflect and fall into the water. Please look at the figure for further clarification (with black, red and green cells representing banks, obstructions and free water respectively). So, if the stone is thrown a distance 7, it will bounce and travel a distance of 3, then finally a distance of 1, having travelled a total distance of 11 (the green path in the figure). If a stone is thrown a distance of 8, it will reach the opposite bank, and if thrown at distances of 2 or 6 it will hit an obstruction during its travel. These are the three red paths in the figure.



You are given a String water. An 'X' represents an obstruction, while a '.' represents water free from obstruction. You are to return an int representing the maximum distance a stone can travel and finally fall in the water, without hitting any obstructions, and without reaching the opposite bank (going beyond the end of the string). You may choose any initial distance for the throw, which starts from the left side of the string. A distance of 1 is the first character of the string, etc. If no initial throw will result in the stone landing in the water without hitting an obstruction, return 0.

Definition

    

Class:

SkipStones

Method:

maxDistance

Parameters:

String

Returns:

int

Method signature:

int maxDistance(String water)

(be sure your method is public)

    

 

 

 

Notes

-

Obstructions are at water level, so the stone will not hit any obstructions while it's in the air.

Constraints

-

water will contain between 1 and 50 elements, inclusive.

-

Each element of water will contain between 1 and 50 characters, inclusive.

-

Each character of each element of water will be 'X' or '.'.

Examples

0)

 

    

"..X.....X..."

Returns: 11

This is the example from the problem statement.

1)

 

    

"...X..."

Returns: 3

If it weren't for the obstruction, we could start with a throw of distance 4, and go a total of 7. But, the best we can do is to throw the stone a distance of 2, and have it skip a distance of 1.

2)

 

    

"....X....X...XXXX.X....."

Returns: 22

12 + 6 + 3 + 1 = 22, is the best case.

3)

 

    

"XXXXXXX.XXX.X.."

Returns: 15

Here, an initial throw of 8 is the only way to avoid hitting an obstruction. Notice that the stone finally falls in the water just before reaching the opposite bank.

 

 

这次的题目可以说并不是太难，也许很多人被全英文的题目给难住了，其实并不应该。像我这个还没过CET4的人都能看得懂，何况是大家呢：）好了，废话不多说了，下面是我写的答案：

public class SkipStones
{
 public int sum;
 public int total;
 public int maxDistance(String water)
 {
  for(int i=water.length();i>0;i--)
  {
   total=0;
   sum=0;
   int j=i;
   do
   {
    sum+=j;
    j/=2;
    }while(j!=0);
   if(sum>water.length()) continue;
   else
   {
    j=i;
    int b=j-1;
    while(j!=0)
    {
     if(water.charAt(b)=='X') break;
     else
     {
      total+=j;
      j/=2;
      b+=j;
      }
     }
    }
    if(total==sum) break;
   }
   if(total==sum) return sum;
   else return 0;
  }
  
  
 public static void main(String[] args)
 {
  SkipStones a=new SkipStones();
  System.out.println("The maxdistance is "+a.maxDistance("..X.....X..."));
  }
 }

FinalFantasy 2005-12-14 16:32 发表评论
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