POJ2499 Binary Tree

Binary Tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5211		Accepted: 2420

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*10⁹) 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

/*
题意：根（a,b），则左孩子为（a+b,b），右孩子为（a,a+b）
    给定（m,n），初试根为（1,1）,从（1,1）到（m,n）需要往左子树走几次，
    往右子树走几次。
思路：逆向思维，从（m,n）到（1,1）。
    给定（m,n），求其父亲，如果m>n，则他父亲是（m-n,n），
    否则（m,n-m）。
代码一：  ----TLE
#include <iostream>

using namespace std;

int main()
{
    int T, a, b, lcnt, rcnt;
    cin >> T;
    for(int i = 1; i <= T; ++i)
    {
        cin >> a >> b;
        lcnt = rcnt = 0;
        while(a != 1 || b != 1)
        {
            if(a >= b)
            {
                ++lcnt;
                a -= b;
            }
            else
            {
                ++rcnt;
                b -= a;
            }
        }
        cout << "Scenario #" << i << ':' << endl;
        cout << lcnt << ' ' << rcnt << endl <<endl;
    }
    return 0;
}

代码二：
用除法代替减法，得到的商即为往左走的次数，最后的m=m%n。
n>m时情况类推。
需要特别注意的是：
    如果m>n,m%n == 0 怎么办？因为根（1,1）不可能有0存在，所以特殊处理一下：
次数：m/n-1；m=1
*/
#include <iostream>

using namespace std;

int main()
{
    int T, a, b, lcnt, rcnt;
    cin >> T;
    for(int i = 1; i <= T; ++i)
    {
        cin >> a >> b;
        lcnt = rcnt = 0;
        while(a != 1 || b != 1)
        {
            if(a >= b)
            {
                if(a % b)
                {
                    lcnt += a/b;
                    a %= b;
                }
                else
                {
                    lcnt += a/b-1;
                    a = 1;
                }
            }
            else
            {
                if(b % a)
                {
                    rcnt += b/a;
                    b %= a;
                }
                else
                {
                    rcnt += b/a-1;
                    b = 1;
                }
            }
        }
        cout << "Scenario #" << i << ':' << endl;
        cout << lcnt << ' ' << rcnt << endl <<endl;
    }
    return 0;
}