poj 2499 binary tree (平衡二叉树)
来源:互联网 发布:pptv网络电视破解版 编辑:程序博客网 时间:2024/05/18 18:20
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:
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?
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.
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
342 13 417 73
Sample Output
Scenario #1:41 0Scenario #2:2 1Scenario #3:4 6
1 /* 2 题意:根(a,b),则左孩子为(a+b,b),右孩子为(a,a+b) 3 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, 4 往右子树走几次。 5 思路:逆向思维,从(m,n)到(1,1)。 6 给定(m,n),求其父亲,如果m>n,则他父亲是(m-n,n), 7 否则(m,n-m)。 8 代码一: ----TLE 9 #include <iostream>10 11 using namespace std;12 13 int main()14 {15 int T, a, b, lcnt, rcnt;16 cin >> T;17 for(int i = 1; i <= T; ++i)18 {19 cin >> a >> b;20 lcnt = rcnt = 0;21 while(a != 1 || b != 1)22 {23 if(a >= b)24 {25 ++lcnt;26 a -= b;27 }28 else29 {30 ++rcnt;31 b -= a;32 }33 }34 cout << "Scenario #" << i << ':' << endl;35 cout << lcnt << ' ' << rcnt << endl <<endl;36 }37 return 0;38 }39 40 代码二:41 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。42 n>m时情况类推。43 需要特别注意的是:44 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:45 次数:m/n-1;m=146 */47 #include <iostream>48 49 using namespace std;50 51 int main()52 {53 int T, a, b, lcnt, rcnt;54 cin >> T;55 for(int i = 1; i <= T; ++i)56 {57 cin >> a >> b;58 lcnt = rcnt = 0;59 while(a != 1 || b != 1)60 {61 if(a >= b)62 {63 if(a % b)64 {65 lcnt += a/b;66 a %= b;67 }68 else69 {70 lcnt += a/b-1;71 a = 1;72 }73 }74 else75 {76 if(b % a)77 {78 rcnt += b/a;79 b %= a;80 }81 else82 {83 rcnt += b/a-1;84 b = 1;85 }86 }87 }88 cout << "Scenario #" << i << ':' << endl;89 cout << lcnt << ' ' << rcnt << endl <<endl;90 }91 return 0;92 }
0 0
- poj 2499 binary tree (平衡二叉树)
- Balanced Binary Tree 二叉平衡树
- 【LeetCode】Balanced Binary Tree,平衡二叉树
- [LeetCode] Balanced Binary Tree 平衡二叉树
- Leetcode Balanced Binary Tree 平衡二叉树
- 平衡二叉树---Balanced Binary Tree
- lintcode balanced-binary-tree 平衡二叉树
- Balanced Binary Tree(平衡二叉树)
- AVL树(平衡二叉树(Balanced Binary Tree))
- 平衡二叉树(Balance Binary Tree) --AVL树
- Balanced Binary Tree 判断是否平衡二叉树 @LeetCode
- 平衡二叉树判定 AVL Balanced Binary Tree
- Balanced Binary Tree 平衡二叉树的检验
- Balanced Binary Tree --判断平衡二叉树(重重)
- LeetCode | Balanced Binary Tree(平衡二叉树)
- Balanced Binary Tree 平衡二叉树的判断
- leetcode Balanced Binary Tree 平衡二叉树判定
- LeetCode 110 Balanced Binary Tree(平衡二叉树)(*)
- 指针学习笔记(1)——运算符*
- 测试功能
- 【PAT】1093. Count PAT's (25)
- JS学习日记-one
- verilog笔试题
- poj 2499 binary tree (平衡二叉树)
- HDU - 2955 Robberies 01背包
- 红黑树实现
- Parallax:视差视图
- Eclipse配置Struts2问题:ClassNotFoundException: org...dispatcher.ng.filter.StrutsPrepareAndExecuteFilter
- VS2010 Windows API 串口编程 (二)
- 李政轩讲核方法kernel Method 视频笔记
- 使用hibernate 5.02
- cheng@Linux之基础--.o .a. so