Binary Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 774 Accepted Submission(s): 450
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：给你一个二叉树，如果当前节点是i，那么左二子是2*i,右儿子是2*i+1,根节点是1，输入n,k，你从根节点开始走每次只能向左下或右下走，每到一个节点，你可以加上或者减去该节点的下标，你可以走k个节点，包括根节点在内，问怎样可以得到k（保证N<=2^k）
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <map>
#include <bitset>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define CT continue
#define SC scanf
bitset<64> bt;
int main()
{
int cas,kk=0;ll n,k;
SC("%d",&cas);
while(cas--)
{
SC("%lld %lld",&n,&k);
printf("Case #%d:
",++kk);
if(n&1)
{
ll w=((1LL<<k)-1-n)/2;
bt=w;
for(int i=0;i<k-1;i++)
{
if(bt[i]&1) printf("%lld -
",(1LL<<i));
else printf("%lld +
",(1LL<<i));
}
printf("%lld +
",(1LL<<(k-1)));
}
else
{
ll w=((1LL<<k)-n)/2;
bt=w;
for(int i=0;i<k-1;i++)
{
if(bt[i]&1) printf("%lld -
",(1LL<<i));
else printf("%lld +
",(1LL<<i));
}
printf("%lld +
",(1LL<<(k-1))+1);
}
}
return 0;
}
可以发现，如果一直沿着左边走k步的话，能表示的最大的数是2^k-1，但是因为N<=2*^k,那么我们可以发现如果最后一步走右边的话，就可以得到2^k,然后再依次减下某个节点。可以发现这两种情况刚好能表示1-2^k
思路：构造出比较极端的例子