Binary Tree

題目連接：

http://acm.hdu.edu.cn/showproblem.php?pid=5573

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 +

Hint

題意

有一顆完全二叉樹，第一個節點是1，他的左兒子就是i x 2，右兒子是i x 2+1

然后讓你找到一個路徑，使得通過加減恰好向下走k步之后，權值和為n

題解：

構造題

首先我們可以分析得到，所有答案都可以通過走 1,2,4,8,16.....來得到

為什么？因為n<=2^k.

如果是奇數，那么我們最后一步就選左兒子，否則選擇右兒子

路徑問題解決了，我們就差符號問題了

符號問題，我們就可以通過找規律來解決

比如 n的二進制為100101

那么我們k步的符號就可以是110010(1表示+,0表示-)

就第一個位置是1，其他位置都是0就好了

比如001,你可以通過4-2-1來構造出來，就是001->100

講得比較玄乎，大家意會吧 T T

代碼

#include<bits/stdc++.h> using namespace std;long long n,k; int flag = 0; int a[100]; int main() {int t;scanf("%d",&t);for(int cas=1;cas<=t;cas++){scanf("%lld%lld",&n,&k);flag = 0;memset(a,0,sizeof(a));if(n%2==0){flag = 1;n = n-1;}int tmp = 0;for(int i=k-1;i>=0;i--){a[i]=1-tmp;tmp=1;if((n>>i)&1)tmp=0;}printf("Case #%d:\n",cas);long long now = 1;for(int i=0;i<k-1;i++){printf("%lld ",now);now = now*2;if(a[i])printf("+\n");else printf("-\n");}if(flag)printf("%lld ",now+1);else printf("%lld ",now);if(a[k-1])printf("+\n");else printf("-\n");} }

轉載于:https://www.cnblogs.com/qscqesze/p/5059326.html

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