A Rational Sequence

最新推荐文章于 2019-08-21 10:10:00 发布

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最新推荐文章于 2019-08-21 10:10:00 发布

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分类专栏：算法和数据结构

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本文介绍了一种基于无限满二叉树的理性数序列生成算法。通过层级遍历方式，实现了从任意给定的理性数找到序列中紧随其后的下一个理性数的功能。文章详细解析了算法的实现思路，并提供了完整的代码示例。

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题目描述

An infinite full binary tree labeled by positive rational numbers is defi ned by:
• The label of the root is 1/1.
• The left child of label p/q is p/(p+q).
• The right child ofl abel p/q is (p+q)/q.
The top of the tree is shown in the following figure:

A rational sequence is defined by doing a level order (breadth first) traversal of the tree (indicated by the light dashed line). So that:
F(1) = 1/1, F(2) = 1/2, F(3) = 2/1, F(4) = 1/3, F(5) = 3/2, F(6) = 2/3, . . .
Write a program which takes as input a rational number, p/q, in lowest terms and fi nds the next rational number in the sequence. That is, if F(n) = p/q, then the result is F(n + 1).

输入

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set should be processed identically and independently.
Each data set consists of a single line of input. It contains the data set number, K, which is then followed by a space, then the numerator of the fraction, p, followed immediately by a fonward slash (/),followed immediately by the denominator of the fraction, q. Both p and q will be relatively prime and 0 ≤ p, q ≤ 2147483647.

输出

For each data set there is a single line of output. It contains the data set number, K, followed by a single space which is then followed by the numerator of the fraction, followed immediately by a forward slash (‘/’) followed immediately by the denominator of the fraction. Inputs will be chosen such that neither the numerator nor the denominator will overfl ow a 32-bit integer.

样例输入

5
1 1/1
2 1/3
3 5/2
4 2178309/1346269
5 1/10000000

样例输出

1 1/2
2 3/2
3 2/5
4 1346269/1860498
5 10000000/9999999

【思路】

最右边的一列，他的下一个就是1/p+1；
是左子树，他的下一个就是q/q-p；
是右子树，就找离他最近的左子树，记录层数差距，再一直找他兄弟的左子树。

【代码】

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{
    int t,k,a,b,p,q;
    char c;
    cin>>t;
    while(t--)
    {
        scanf("%d %d/%d",&k,&a,&b);
        if(a<b)
        {
            p=b;
            q=b-a;
        }
        else if(b==1)
        {
            p=1;
            q=a+1;
        }
        else
        {
            int ans=a/b;
            a=a-b*ans;
            p=b;
            q=b-a+p*ans;
        }
        cout<<k<<" "<<p<<"/"<<q<<endl;
    }
    return 0;
}