Binary Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 251 Accepted Submission(s): 143
Special Judge
Problem Description
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 1. Say froot=1.
And for each node u, labels as fu, the left child is fu×2 and right child is fu×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 N years, only if he could collect exactly N soul 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 x, the number at the node is fx (remember froot=1), he can choose to increase his number of soul gem by fx, or decrease it by fx.
He will walk from the root, visit exactly K nodes (including the root), and do the increasement or decreasement as told. If at last the number is N, 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 N, K, help the King find a way to collect exactly N soul gems by visiting exactly K nodes.
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 1. Say froot=1.
And for each node u, labels as fu, the left child is fu×2 and right child is fu×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 N years, only if he could collect exactly N soul 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 x, the number at the node is fx (remember froot=1), he can choose to increase his number of soul gem by fx, or decrease it by fx.
He will walk from the root, visit exactly K nodes (including the root), and do the increasement or decreasement as told. If at last the number is N, 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 N, K, help the King find a way to collect exactly N soul gems by visiting exactly K nodes.
Input
First line contains an integer T,
which indicates the number of test cases.
Every test case contains two integers N and K, which indicates soul gems the frog king want to collect and number of nodes he can visit.
⋅ 1≤T≤100.
⋅ 1≤N≤109.
⋅ N≤2K≤260.
Every test case contains two integers N and K, which indicates soul gems the frog king want to collect and number of nodes he can visit.
⋅ 1≤T≤100.
⋅ 1≤N≤109.
⋅ N≤2K≤260.
Output
For every test case, you should output "Case #x:" first, where x indicates
the case number and counts from 1.
Then K lines follows, each line is formated as 'a b', where a is node label of the node the frog visited, and b is either '+' or '-' which means he increases / decreases his number by a.
It's guaranteed that there are at least one solution and if there are more than one solutions, you can output any of them.
Then K lines follows, each line is formated as 'a b', where a is node label of the node the frog visited, and b is either '+' or '-' which means he increases / decreases his number by a.
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
2
5 3
10 4
Sample Output
Case #1:
1 +
3 -
7 +
Case #2:
1 +
3 +
6 -
12 +
Source
题意：
给你一个n和k，要求用k层的完全二叉树，从根节点走到叶子节点，然后用走过的这k个数的加减组成n，多组解
输出一个结果即可
/*
开始实在是没想到什么方法，于是搜了一发GG.
后来看题解才发现N≤2K≤2^60这个条件很重要- -
我们只需要考虑二叉树最左边那条边，1 2 4 8 16 ......（好机智）
于是愉快地的开始写了
根据奇偶的不同再看是取最底端的左孩子或右孩子
对于左边所有节点求和t，dis = t-n能求出我们比预计多了多少
然后只要不加上dis/2即可
而且我们发现最左边 2^0,2^1,2^2.... ,所有把dis转换成2进制，有1的地方取‘-’即可
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int vis[10005];
int n,k;
int main()
{
int T;
int cas =1;
scanf("%d",&T);
while(T--)
{
printf("Case #%d:
",cas++);
scanf("%d%d",&n,&k);
int t;
if(n % 2 == 0)
t = 1<<k ;
else
t = (1<<k) -1;
int dis = (t - n)/2;
memset(vis,0,sizeof(vis));
for(int i = k; i >= 0; i--)
{
if(1 & (dis >> i))
vis[i+1] = 1;
}
int cur = 1;
for(int i = 1; i < k; i++)
{
if(vis[i])
printf("%d %c
",cur,'-');
else
printf("%d %c
",cur,'+');
cur *= 2;
}
printf("%d %c
",n%2 ? 1<<(k-1):(1<<(k-1))+1,vis[k] ? '-':'+');
}
return 0;
}