版权声明：抱最大的希望，为最大的努力，做最坏的打算。 https://blog.csdn.net/qq_37748451/article/details/86497367
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 +
Source
2015ACM/ICPC亚洲区上海站-重现赛（感谢华东理工）规律构造
题意：从1走到第k层，下一层的数是上一层的数*2或者*2+1，可以选择加上或者减去走的数，最终要求结果为n
输出每层走的数，和是加还是减。
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-6
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define N 300001
using namespace std;
#define ll long long int
long long n,m;
int main()
{
int t;
scanf("%d",&t);
for(int i=1;i<=t;i++)
{
scanf("%lld%lld",&n,&m);
printf("Case #%d:\n",i);
bool flag=1;
if(n%2!=0)
{
n=n/2+1;
flag=0;
}
else
n=n/2;
ll fg=(1<<(m-1))-n;
ll tmp=1;
for(int i=0;i<m-1;i++)
{
printf("%lld ",tmp);
if(fg%2!=0)
printf("-\n");
else
printf("+\n");
fg>>=1;
tmp<<=1;
}
if(flag)
tmp++;
printf("%lld ",tmp);
if(fg%2!=0)
printf("-\n");
else
printf("+\n");
}
return 0;
}