Written by razrlele
23:51 August 16, 2014
Binary Tree
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6424 | Accepted: 2954 |
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:
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
|
1 2 3 4 |
3 42 1 3 4 17 73 |
Sample Output
|
1 2 3 4 5 6 7 8 |
Scenario #1: 41 0 Scenario #2: 2 1 Scenario #3: 4 6 |
Source
TUD Programming Contest 2005 (Training Session), Darmstadt, Germany
这道题首先得明白, (x, y)中如果x>y那么它的父亲则是(x-y， y), 否则是(x, y-x), 这样就得出了下面的代码:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 |
/************************************************************************* > File Name: poj2499.cpp > Author: razrLeLe > Personal homepage: https://razrLeLe.com > Created Time: Sat 16 Aug 2014 09:30:48 PM CST ************************************************************************/ #include<iostream> using namespace std; #include<cstdio> int main() { int casenum; scanf("%d", &casenum); int x, y; int countl, countr; for(int i = 1; i <= casenum; i++) { countl = 0; countr = 0; scanf("%d%d", &x, &y); while(x!=1 || y!=1) { if(x>y) { x = x-y; countl++; } else { y = y-x; countr++; } } printf("Scenario #%d:n%d %dnn", i, countl, countr); } } |
咳咳, 题目到这里显然还没有完, 即使这个题是这么的水, 也不是那么一下两下可以搞定的, 所以只是上面的代码必然超时.
这里超时的原因肯定是当x>y的时候x的值如果只是减一减不够快, 所以这里改成取余就可以了, 这里要注意一下xy中有一个先到一的情况
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 |
/************************************************************************* > File Name: poj2499.cpp > Author: razrLeLe > Personal homepage: https://razrLeLe.com > Created Time: Sat 16 Aug 2014 10:32:34 PM CST ************************************************************************/ #include<iostream> using namespace std; #include<cstdio> int main() { int casenum; scanf("%d", &casenum); int x, y; int countl, countr; for(int i = 1; i <= casenum; i++) { countl = 0; countr = 0; scanf("%d%d", &x, &y); while(x!=1 || y!=1) { if(x>y) { if(y == 1) { countl += x-1; x = 1; } else { countl += x/y; x %= y; } } else { if(x == 1) { countr += y-1; y = 1; } else { countr += y/x; y %= x; } } } printf("Scenario #%d:n%d %dnn", i, countl, countr); } } |