poj2499--Binary Tree

Binary Tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5832		Accepted: 2705

Description

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?

Input

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*10 ⁹) that represent 
a node (i, j). You can assume that this is a valid node in the binary tree described above.

Output

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.

Sample Input

3
42 1
3 4
17 73

Sample Output

Scenario #1:
41 0

Scenario #2:
2 1

Scenario #3:
4 6

Source

TUD Programming Contest 2005 (Training Session), Darmstadt, Germany

一开始没读懂意思，以为是个二叉树。。。。。。。。

后来才知道可以倒着推回去，其中要用n/m 一类的可以减少时间

#include <stdio.h>
int main()
{
    int t , tt ;
    scanf("%d", &t);
    for(tt = 1 ; tt <= t ; tt++)
    {
        int n , m , l = 0 , r = 0 ;
        scanf("%d %d", &n, &m);
        while(1)
        {
            if(n == 1 && m == 1)
                break;
            if(n > m)
            {
                if(m == 1)
                {
                    l += (n-1) ;
                    n = 1 ;
                }
                else
                {
                    l += n / m ;
                    n = n % m ;
                }
            }
            else
            {
                if(n == 1)
                {
                    r += (m-1) ;
                    m = 1 ;
                }
                else
                {
                    r += m /n ;
                    m = m % n ;
                }
            }
        }
        printf("Scenario #%d:\n%d %d\n\n", tt, l , r);
    }
    return 0;
}