【HDU 5573】Binary Tree

The Old Frog King lives on the root of an infinite tree. According to the law, each node should connect to exactly two nodes on the next level, forming a full binary tree. 

Since the king is professional in math, he sets a number to each node. Specifically, the root of the tree, where the King lives, is 11. Say froot=1froot=1. 

And for each node uu, labels as fufu, the left child is fu×2fu×2 and right child is fu×2+1fu×2+1. The king looks at his tree kingdom, and feels satisfied. 

Time flies, and the frog king gets sick. According to the old dark magic, there is a way for the king to live for another NN years, only if he could collect exactly NNsoul gems. 

Initially the king has zero soul gems, and he is now at the root. He will walk down, choosing left or right child to continue. Each time at node xx, the number at the node is fxfx (remember froot=1froot=1), he can choose to increase his number of soul gem by fxfx, or decrease it by fxfx. 

He will walk from the root, visit exactly KK nodes (including the root), and do the increasement or decreasement as told. If at last the number is NN, then he will succeed. 

Noting as the soul gem is some kind of magic, the number of soul gems the king has could be negative. 

Given NN, KK, help the King find a way to collect exactly NN soul gems by visiting exactly KK nodes.

Input

First line contains an integer TT, which indicates the number of test cases. 

Every test case contains two integers NN and KK, which indicates soul gems the frog king want to collect and number of nodes he can visit. 

⋅⋅ 1≤T≤1001≤T≤100. 

⋅⋅ 1≤N≤1091≤N≤109. 

⋅⋅ N≤2K≤260N≤2K≤260.

Output

For every test case, you should output " Case #x:" first, where xx indicates the case number and counts from 11. 

Then KK lines follows, each line is formated as 'a b', where aa is node label of the node the frog visited, and bb is either '+' or '-' which means he increases / decreases his number by aa. 

It's guaranteed that there are at least one solution and if there are more than one solutions, you can output any of them. 

Sample Input

25 310 4

Sample Output

Case #1:1 +3 -7 +Case #2:1 +3 +6 -12 +

1， 2， 4， 2^k可以构造出所有小于2^（k + 1）的数，

本来想构造二叉树来解题，树都建好了，大佬忽然告诉我这道题不需要建树，再看一下数据，太大，哪怕剪枝都要超时，百度了一下发现是二进制的构造问题

求出sum后，根据sum - n来确定最后一个数取2^k或2^k + 1,最后根据x的二进制判断正负号

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<stdio.h>#include<vector>using namespace std;typedef long long ll;ll N, ans;int K, sum, x;int main(){    int T, k = 1;    scanf("%d", &T);    while(T--)    {        scanf("%I64d %d", &N, &K);        cout<<"Case #"<<k++<<": "<<endl;        if(N & 1)        {            sum = (1 << K) - 1;             //(2 ^ K) - 1            x = (sum - N) / 2;            for(int i = 0; i < K; i++)            {                cout<<(1 << i)<<" ";                if(x & 1)                    cout<<"-"<<endl;                else                    cout<<"+"<<endl;                x >>= 1;            }        }        else        {            sum = (1 << K);            x = (sum - N) / 2;            for(int i = 0; i < K - 1; i++)            {                cout<<(1 << i)<<" ";                if(x & 1)                    cout<<"-"<<endl;                else                    cout<<"+"<<endl;                x >>= 1;            }            cout<<(1 << K - 1) + 1<<" ";            if(x & 1)                cout<<"-"<<endl;            else                cout<<"+"<<endl;        }    }    return 0;}