hnust 1450 Bird tree

1450: Bird tree

时间限制: 1 Sec   内存限制: 64 MB
提交: 41   解决: 14
[ 提交][ 状态][ 讨论版]

题目描述

The Bird tree is an infinite binary tree, whose first 5 levels look as follows:

It can be defined as follows:

This is a co-recursive definition in which both occurrences of bird refer to the full (infinite) tree. The expression bird + 1 means that 1 is added to every fraction in the tree, and 1∕bird means that every fraction in the tree is inverted (so ^a∕_b becomes ^b∕_a).

Surprisingly, the tree contains every positive rational number exactly once, so every reduced fraction is at a unique place in the tree. Hence, we can also describe a rational number by giving directions (L for left subtree, R for right subtree) in the Bird tree. For example, ²∕₅ is represented by LRR. Given a reduced fraction, return a string consisting of L’s and R’s: the directions to locate this fraction from the top of the tree.

输入

On the first line a positive integer: the number of test cases, at most 100. After that per test case:

• one line with two integers a and b (1 ≤ a,b ≤ 10⁹), separated by a ’/’. These represent the numerator and denominator of a reduced fraction. The integers a and b are not both equal to 1, and they satisfy gcd(a,b) = 1.

For every test case the length of the string with directions will be at most 10 000.

输出

Per test case:

• one line with the string representation of the location of this fraction in the Bird tree.

样例输入

31/22/57/3

样例输出

LLRRRLLR

报告人：SpringWater（GHQ）

通项公式：

 father[1……2^N] = b/a;
leftchild[1……2^N] = a/(a + b);
rightchild[1……2^N] = (a + b) /b; 

代码：

#include<stdio.h>
int main()
{
    int a,b,cas,t;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d/%d",&a,&b);
        while(!(a==1&&b==1))
           if(a>b)putchar('R'),t=a,a=b,b=t-b;
           else putchar('L'),t=a,a=b-a,b=t;
        putchar('\n');
    }
    return 0;
}