hdu5573 Binary Tree 二進制

Binary Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 952    Accepted Submission(s): 558
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 Nsoul 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 +

思路參考：http://blog.csdn.net/u013068502/article/details/50094561

有兩個地方想了一會纔想明白，一個是d爲什麼要除以2，因爲a + b + c要變成a - b - c那麼也就是

a + b + c - 2(b + c)，所以要變號的話肯定是(a + b + c - (a - b - c)) / 2 = b + c，然後再把b和c提取出來變號就可以了

還有一個地方是最後加減1的問題，爲什麼d必須是偶數，因爲最小變換是1變號到-1，1 - (-1) = 2，也就是說最小

變換相差還是偶數呢，如果d = 100111(二進制)，你尾巴上的那個1怎麼辦a - (-a) = 1，2a = 1，a = 1/2，肯定不行的

所以d必須得是偶數，例如d = 10(二進制)，a - (-a) = 2a = 2，a = 1，只需要把1變號成-1就可以了，是可以滿足的

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <map>

using namespace std;

typedef long long ll;

int T, n, k, kase;
ll sum, d, x, t;
bool flag;
map<ll, int> mp;

int main()
{
	cin >> T;
	kase = 0;
	while (T--) {
		scanf("%d%d", &n, &k);
		sum = 0;
		for (int i = 0; i < k; i++) {
			sum = sum + ((ll)1 << i);
		}
		flag = false;
		d = sum - n;
		d = (d % 2 ? flag = true, d + 1 : d);
		x = d >> 1;
		mp.clear();
		for (int i = 0; i < 61; i++) {
			t = ((ll)1 << i) & x;
			if (t) {
				mp[t] = 1;
			}
		}
		printf("Case #%d:\n", ++kase);
		for (int i = 0; i < k; i++) {
			t = ((ll)1 << i);
			if (flag && i == k - 1) {
				t++;
			}
			printf("%lld ", t);
			if (mp[t] == 1) {
				printf("-");
			}
			else {
				printf("+");
			}
			puts("");
		}
	}
	return 0;
}