hdu 5573 Binary Tree【思维+递推】

Binary Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 507    Accepted Submission(s): 292
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 isfx (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 +

Source

2015ACM/ICPC亚洲区上海站-重现赛（感谢华东理工）

 

题目大意：

有一个二叉树，根节点编号为1，其子节点编号为2x和2x+1，如果到了某一个点，那么对应可以减少这个点的标号的数值，也可以增加这个点的标号的数值，给你一个数值和走的步数，问你要一个合理的方案并且输出。

思路:

1、一开始想到将问题转化到一直向2*x这个方向方向出发，对于这个方向上的数值进行+或者-的操作。这样就能从发散的去寻找解变成二进制求解。

2、然后为了验证思路的正确性，暴力敲了发深搜的，验证出：确实无论什么数据，如果合法，那么一定可以由这个方向（2*x）的数完成这个任务。然后又多敲了几发数据，发现如果输入的是偶数，那么沿着2*x方向可以一直走下去，但是最后一步需要+1.

3、那么问题就转化到了这样：

沿着这样的一条路一直走下去，对应每个数取正值或者负值，累加。

1、2、4、8、16、......................

4、然后问题转化到二进制数上来：将一个数转化成二进制数，0表示减法，1表示加法。

5、递推找到规律：将n先用1<<k之后，再减去n得到值tmp：1<<k-n;然后将tmp的二进制数求出来。然后从右向左第二位开始判断，如果是1，表示这位数值为负值，否则为正。比如：

7 5

tmp=1<<5-n=25

9的二进制数：11001

那么输出为：

1 +

2 +

4 -

8 -

16 +

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll long long int
int main()
{
    int t;
    int kase=0;
    scanf("%d",&t);
    while(t--)
    {
        ll n,k;
        scanf("%I64d%I64d",&n,&k);
        ll tmp=(1ll<<k);
        tmp-=n;
        printf("Case #%d:\n",++kase);
        for(int i=1;i<k;i++)
        {
            if(tmp&(1<<i))printf("%I64d -\n",(1ll<<(i-1)));
            else printf("%I64d +\n",(1ll<<(i-1)));
        }
        tmp=(1ll<<(k-1));
        if(n%2==0)tmp++;
        printf("%I64d +\n",tmp);
    }
}