Binary Tree 分类： POJ 2015-06-12 20:34 17人阅读 评论(0) 收藏

Binary Tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6355		Accepted: 2922

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

给一个节点，计算从根节点到所给节点的所经过的左节点，和右节点的个数。

#include <cstdio>
#include <string.h>
#include <cmath>
#include <iostream>
#include <algorithm>
#define WW freopen("output.txt","w",stdout)
using namespace std;
int main()
{
    int T,w=1;
    int L,R;
    int sum;
    int ans;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&L,&R);
        sum=0;
        ans=0;
        while(L!=1||R!=1)
        {
            if(L>R)
            {
                int a=(L-1)/R;//进行优化不然会TLE
                ans+=a;
                L-=(a*R);
            }
            else if(R>L)
            {
                int a=(R-1)/L;
                sum+=a;
                R-=(L*a);
            }

        }
        printf("Scenario #%d:\n",w++);
        printf("%d %d\n\n",ans,sum);//两个换行
    }
    return 0;
}

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