Binary Tree（二叉树+思维）

Binary Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 138    Accepted Submission(s): 73 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.

 

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.

 

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.

 

Sample Input

2 5 3 10 4

 

Sample Output

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

 

题解：

给你一个满二叉树，从根开始往下走，每次加上或减去当前节点的权值，最终结果等与N，2^k>=N，由于1+2+4+....+2^(k-1)等于2^k-1（前k层的和）所以只需要减去x=2^k-1-N就好了，而x可以表示为2^a+2^b+2^c......所以用lowbit标记就好了；

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define T_T while(T--)
#define P_ printf(" ")
#define mem(x,y) memset(x,y,sizeof(x))
typedef long long LL;
LL lowbit(LL x){return x&(-x);}
LL sign[65];
int main(){
	int kase=0;
	int T,K;
	LL N;
	SI(T);
	T_T{
		mem(sign,0);
		SL(N);SI(K);
		LL x=(1<<K)-N-1;
	//	PL(x);puts("");
		if(x&1)x++;
		//printf("%lld
",x);
		//x>>=1;
		x/=2;
		while(x>0){
			LL temp=lowbit(x);
			x-=temp;
			int p=0;
			while(temp>0){
				p++;
				temp>>=1;
			}
			sign[p]=1;
		}
		printf("Case #%d:
",++kase);
		for(int i=1;i<=K;i++){
			if(i==K){
				if(((1<<K)-N-1)&1)PL((1<<(i-1))+1),P_;
				else PL(1<<(i-1)),P_;
			}
			else PL(1<<(i-1)),P_;
			if(sign[i])puts("-");
			else puts("+");
		}
	}
	return 0;
}