POJ-2499 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*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

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Sample Output

Scenario #1:
41 0

Scenario #2:
2 1

Scenario #3:
4 6

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 1 /*
 2  The question ： root （a,b）, The left child is （a+b,b）, The right child is （a,a+b）
 3      Given （m,n）, The initial root is （1,1）, from （1,1） To （m,n） You need to walk to the left sub tree several times ,
 4      Walk to the right sub tree several times .
 5  Ideas ： Converse thinking , from （m,n） To （1,1）.
 6      Given （m,n）, Beg his father , If m>n, Then his father is （m-n,n）,
 7      otherwise （m,n-m）.
 8  Code 1 ：  ----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             else
29             {
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  Code 2 ：
41  Use division instead of subtraction , The quotient obtained is the number of times to go left , final m=m%n.
42 n>m And so on .
43  Here's the thing to watch out for ：
44      If m>n,m%n == 0  What do I do ？ Because root （1,1） There can be no 0 There is , So special treatment ：
45  frequency ：m/n-1;m=1
46 */
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                 else
69                 {
70                     lcnt += a/b-1;
71                     a = 1;
72                 }
73             }
74             else
75             {
76                 if(b % a)
77                 {
78                     rcnt += b/a;
79                     b %= a;
80                 }
81                 else
82                 {
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 }

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