【规律】A Rational Sequence

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

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【题解】

　　题意就是让大家根据这颗树构造来跑到下一个点。大家通过观察可以看到，除了最右侧的那个节点外，其他情况都是往上爬树，然后又往下爬。

面对像图  3/2 -> 2/3 这种情况来讲：

1、其实是往上爬，也就是 分子在减少，分母不变。后来发现，一直减去分母直到无法减为止，其实也就是相当于取余运算。

2、然后往右翻一下。分子变成原来的分母，分母变成原来的分母减分子。

3、然后在第一步取余后不是获取了层数吗？然后直接利用层数对分子进行加上分母×层数。

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【代码】

异常简单，如果发现规律的话。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    int T,kase;
    ll p,q;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&kase);
        scanf("%lld/%lld",&p,&q);
        if( q == 1 ){
            printf("%d %d/%lld
",kase,1,p+1);
        }else if( p < q ){
            ll tmp = q;
            q = tmp - p;
            p = tmp ;
            printf("%d %lld/%lld
",kase,p,q);
        }else{
            ll dep = 0 ;
            dep = p/q;
 
            p = p%q;
            //printf("Dep : %d
",dep);
            ll tmp = q;
            q = tmp - p;
            p = tmp ;
 
            //printf("%d %d + %d
",p,q,dep*p);
 
            q = q + dep*p;
            printf("%d %lld/%lld
",kase,p,q);
        }
    }
    return 0 ;
}