ACdream 1187——Rational Number Tree（二进制，递归）

题目：

Description

Consider an infinite complete binary tree where the root node is 1/1 and left and right childs of node p/q are p/(p+q) and (p+q)/q, respectively. This tree looks like:

         1/1    ______|______    |           |   1/2         2/1 ___|___     ___|___ |     |     |     |1/3   3/2   2/3   3/1...

It is known that every positive rational number appears exactly once in this tree. A level-order traversal of the tree results in the following array:

1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, ...

Please solve the following two questions:

1. Find the n-th element of the array, where n starts from 1. For example, for the input 2, the correct output is 1/2.
2. Given p/q, find its position in the array. As an example, the input 1/2 results in the output 2.

Input

The first line of the input gives the number of test cases, T(1 ≤ T ≤ 100).

T test cases follow. Each test case consists of one line.

The line contains a problem id (1 or 2) and one or two additional integers:

1. If the problem id is 1, then only one integer n is given, and you are expected to find the n-th element of the array.
2. If the problem id is 2, then two integers p and q are given, and you are expected to find the position of p/q in the array.

p and q are relatively prime.

1 ≤ n, p, q ≤ 2^64-1

p/q is an element in a tree with level number ≤ 64.

Output

For each test case:

1. If the problem id is 1, then output one line containing "Case #x: p q", where x is the case number (starting from 1), and p, q are numerator and denominator of the asked array element, respectively.
2. If the problem id is 2, then output one line containing "Case #x: n", where x is the case number (starting from 1), and n is the position of the given number.

Sample Input

41 22 1 21 52 3 2

Sample Output

Case #1: 1 2Case #2: 2Case #3: 3 2Case #4: 5

题意：一棵树，每个节点是一个分数。做查询，通过编号确定分数，通过分数确定编号。

思路：二叉树，编号可以用二进制表示。然后逐步递归回去。两种查询要用不同的递归方法但思路基本一致。注意数据范围。要用llu，或者用cin和cout。

代码如下：

#include <iostream>#include <cmath>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <algorithm>#include <string>using namespace std;#define MAXN 1e30#define INF 1e9+7#define MODE 1000000unsigned long long a=1,b=1;void solve_p_q(unsigned long long n){if(n==1){a=b=1;}else if(n&1){solve_p_q((n-1)>>1);a=a+b;}else{solve_p_q(n>>1);b=a+b;}}unsigned long long res=0;void solve_n(unsigned long long p,unsigned long long q){if(p==q)res=1;else if(p>q){solve_n(p-q,q);res=res*2+1;}else{solve_n(p,q-p);res=res*2;}}int main(){    int t;    cin>>t;    int s=1;    for(int s=1;s<=t;s++)    {        int m;        cin>>m;        if(m==1)        {            unsigned long long n;            cin>>n;            solve_p_q(n);            cout<<"Case #"<<s<<": "<<a<<" "<<b<<endl;        }        if(m==2)        {            unsigned long long p,q;            cin>>p>>q;            solve_n(p,q);            printf("Case #%d: ",s);            cout<<res<<endl;        }    }}