The Stern-Brocot tree is an infinite complete binary tree in which the vertices correspond one-for-one to the positive rational numbers, whose values are ordered from the left to the right as in a search tree.

Figure 1 shows a part of the Stern-Brocot tree, which has the first 4 rows. Each node in the tree is marked in a red cycle. The value in the node is the mediant of the left and right fractions. The mediant of two fractions A/B and C/D is defined as (A+C)/(B+D).

To construct the Stern-Brocot tree, we first define the left fraction of the root node is 0/1, and the right fraction of the root node is 1/0. So the value in the root node is the mediant of 0/1 and 1/0, which is (0+1)/(1+0)=1/1. Then the value of root node becomes the right fraction of the left child, and the left fraction of the right child. For example, the 1st node in row2 has 0/1 as its left fraction and 1/1(which is the value of its parent node) as its right fraction. So the value of the 1st node in row2 is (0+1)/(1+1)=1/2. For the same reason, the value of the 2nd node in row2 is (1+1)/(1+0)=2/1. This construction progress goes on infinitly. As a result, every positive rational number can be found on the Stern-Brocot tree, and can be found only once.

Given a rational number in form of P/Q, find the position of P/Q in the Stern-Brocot Tree.

輸入描述:

Input consists of two integers, P and Q (1<=P,Q<=1000), which represent the rational number P/Q. We promise P and Q are relatively prime.

輸出描述：

Output consists of two integers, R and C.
R indicates the row index of P/Q in the Stern-Brocot Tree, C indicates the index of P/Q in the row.

Both R and C are base 1.
We promise the position of P/Q is always in the first 12 rows of the Stern-Brocot tree, which means R<=12.

輸入例子1:

5 3

輸出例子1:

4 6

思路：

在這裡我嘗試用遞迴的方式來實現此功能。

具體而言，要定義一個尋找目標分數位置的函式search：

search(P,Q,X,Y,Row,Col,Lx,Ly,Rx,Ry)

這裡的P/Q為要尋找的目標分數，X/Y為 當前已經找到的分數，Row與Col分別為當前找到的分數X/Y所在的位置，Lx/Ly,Rx/Ry分別為要想找到X/Y所用的兩個數，即X =Lx+Rx，Y =Ly+Ry。

當向左搜尋時:

X(i+1)=X(i)+Lx(i)，Y(i+1)=Y(i)+Ly(i)，Row(i+1)=Row(i)+1，Col(i+1)=2*Col(i)-1，Lx(i+1)=Lx(i)，Ly(i+1)=Ly(i)，Rx(i+1)=X(i)，Ry(i+1)=Y(i)

當向右搜尋時:

X(i+1)=X(i)+Rx(i)，Y(i+1)=Y(i)+Ry(i)，Row(i+1)=Row(i)+1，Col(i+1)=2*Col(i)，Lx(i+1)=X(i)，Ly(i+1)=Y(i)，Rx(i+1)=Rx(i)，Ry(i+1)=Ry(i)

由此我們可以寫出程式碼如下：

def search(P,Q,X,Y,Row,Col,Lx,Ly,Rx,Ry):
    #終止條件
    if P==X and Q==Y:
        return Row,Col

    #向左搜尋
    if P/Q < X/Y:
        r,c = search(P,Q,X+Lx,Y+Ly,Row+1,2*Col-1,Lx,Ly,X,Y)
        return r,c
    #向右搜尋
    if P/Q > X/Y:
        r,c = search(P,Q,X+Rx,Y+Ry,Row+1,2*Col,X,Y,Rx,Ry)
        return r,c

getNumFromString = lambda x:list(map(int,x.strip().split()))
p,q = getNumFromString(input())
R,C = search(p,q,1,1,1,1,0,1,1,0)
print(R,C)

通過看別人寫的程式碼，真的感慨自己跟人家的差距真的太大了！

唯有多看多學才能一點一點進步，好好加油！！！