HDU 5573 Binary Tree （构造）

Binary Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1353    Accepted Submission(s): 796
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 +

 

一开始基于本题的性质，也能想到跟二进制有关，但是没想到具体的运用方法.- -

当时注意到该题的范围说明很特别，但没特别在意，这是很不应该的.

正确的思考路线，我想可以先打个表，就会发现，这些数据很多的合理组成方式都跟1 2 4 8等等密切相关，

然后将思路往二进制上面转化，把1个数转化为2进制之后，就会是1+2+4等类似的形式，比如101就是1+4，

但是这里的二进制是不仅不加，还要减去对应的位数，也就是说对于2的k次方-1来说，就会减去两倍的位数.

那么我们将2的k次方-1和n的差，除以2，再按2进制进行题目要求的操作，是不是就能得到n了呢。

如果差为奇数，我们将差先加一再除以2,这样会多减1，所以最后一个数走右边即可。

上面讨论的是n<2的k次方的情况，最后特殊一下n==2的k次方的情况

代码如下：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
ll a[110];
int main()
{
   ll t,n,k,Case=0,cha,tmp,cnt,h;
   scanf("%lld",&t);
   while(t--)
   {
       Case++;
     scanf("%lld%lld",&n,&k);
        cha=(1<<(k))-1-n;
     memset(a,0,sizeof(a));
       if(cha%2==1)
       tmp=(cha+1)/2;
       else
       tmp=cha/2;
       cnt=0;
       while(tmp>0)
       {
          a[cnt++]=tmp%2;
          tmp=tmp/2;
       }
       printf("Case #%lld:\n",Case);
       for(ll i=0;i<k-1;i++)
       {
           h=(1<<i);
         if(a[i]==0)
         {
          printf("%lld +\n",h);
         }
         else
         printf("%lld -\n",h);
       }
      if(cha%2==1||n==(1<<(k)))
       printf("%lld +\n",(1<<k-1)+1);
     else
        printf("%lld +\n",(1<<k-1));

   }
    return 0;
}