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2013
11-11

POJ 2499 Binary Tree [解题报告] Java

Binary Tree

问题描述 :

Background

Binary trees are a common data structure in computer science. In this problem we will look at an infinite binary tree where the nodes contain a pair of integers. The tree is constructed like this:
  • The root contains the pair (1, 1).
  • If a node contains (a, b) then its left child contains (a + b, b) and its right child (a, a + b)

Problem

Given the contents (a, b) of some node of the binary tree described above, suppose you are walking from the root of the tree to the given node along the shortest possible path. Can you find out how often you have to go to a left child and how often to a right child?

输入:

The first line contains the number of scenarios.

Every scenario consists of a single line containing two integers i and j (1 <= i, j <= 2*109) that represent

a node (i, j). You can assume that this is a valid node in the binary tree described above.

输出:

The output for every scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario starting at 1. Then print a single line containing two numbers l and r separated by a single space, where l is how often you have to go left and r is how often you have to go right when traversing the tree from the root to the node given in the input. Print an empty line after every scenario.

样例输入:

3
42 1
3 4
17 73

样例输出:

Scenario #1:
41 0

Scenario #2:
2 1

Scenario #3:
4 6

解题代码:

//* @author: [email protected]
import java.util.*;
public class Main
{
 public static void main(String[] args)
 {
  Scanner in=new Scanner(System.in);
  int a=in.nextInt();
  int cnt=0;
  while((a--)!=0)
  {
   int b=in.nextInt();
   int c=in.nextInt();
   cnt++;
   int d=0,e=0;
   while(true)
   {
    if(b>c)
    {
     d+=b/c;
     b=b%c;
     if(b==0) break;
    }
    else
    {
	e+=c/b;
	c=c%b;
	if(c==0) break;
     }
    }
    if(b==0) d--;
    if(c==0) e--;
    System.out.println("Scenario #"+cnt+":");
    System.out.println(d+" "+e);
    System.out.println();
   }
 }
}

  1. 第23行:
    hash = -1是否应该改成hash[s ] = -1

    因为是要把从字符串s的start位到当前位在hash中重置

    修改提交后能accept,但是不修改居然也能accept

  2. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。