2014计算机学科夏令营上机考试H:Binary Tree 特殊二叉树自底向上求解（找规律数学题）

总时间限制: 1000ms 内存限制: 65536kB题目
描述
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?
输入
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.
输出
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.
样例输入

3
42 1
3 4
17 73

样例输出

Scenario #1:
41 0

Scenario #2:
2 1

Scenario #3:
4 6

思路：乍看像是要建树，实际上是找规律的数学题，最好把树画出来然后自底向上求解
注意存在倍数关系，用除法代替减法，否则会超时。
代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<queue>
#include<set>
#include<stack>
using namespace std;
int n,a,b;
int l=0,r=0;
void solve(int a,int b)
{
    while(a!=1||b!=1)
    {
        if(a==1)
        {
            r+=(b-1);
            return ;
        }
        if(b==1)
        {
            l+=(a-1);
            return;
        }
        if(a>b)
        {
            l+=a/b;
            a=a%b;
        }
        else if(a<b)
        {
            r+=b/a;
            b=b%a;
        }
    }
}
int main()
{
    int label=1;
    //freopen("input.txt","r",stdin);
    while(cin>>n)
    {
        for(int i=0;i<n;i++)
        {
            l=0;r=0;
            cin>>a>>b;
            solve(a,b);
            cout<<"Scenario #"<<label++<<":"<<endl;
            cout<<l<<" "<<r<<endl<<endl;
        }
    }
    return 0;
}