Phoenix and Science

题意翻译

小P觉得很无聊于是开始繁殖细菌。 在第一天，有一个质量为1的细菌。 接下来的每一天，会有若干个细菌分裂成两个（0$\le$分裂的细菌的数量$\le$当前的细菌数），两个细菌的质量皆为原来细菌的一半。 所有细菌分裂完之后每个细菌的质量增长1。 给出你一个整数$n$，如果没有方案能够在若干天后使得细菌质量的和等于$n$，输出`-1`，否则输出两行，第一行是最少花费的天数，第二行是每一天分裂的细菌的数量。 translate by luogu@$\text{longer\_name}$

题目描述

Phoenix has decided to become a scientist! He is currently investigating the growth of bacteria. Initially, on day $ 1 $ , there is one bacterium with mass $ 1 $ . Every day, some number of bacteria will split (possibly zero or all). When a bacterium of mass $ m $ splits, it becomes two bacteria of mass $ \frac{m}{2} $ each. For example, a bacterium of mass $ 3 $ can split into two bacteria of mass $ 1.5 $ . Also, every night, the mass of every bacteria will increase by one. Phoenix is wondering if it is possible for the total mass of all the bacteria to be exactly $ n $ . If it is possible, he is interested in the way to obtain that mass using the minimum possible number of nights. Help him become the best scientist!

输入输出格式

输入格式

The input consists of multiple test cases. The first line contains an integer $ t $ ( $ 1 \le t \le 1000 $ ) — the number of test cases. The first line of each test case contains an integer $ n $ ( $ 2 \le n \le 10^9 $ ) — the sum of bacteria masses that Phoenix is interested in.

输出格式

For each test case, if there is no way for the bacteria to exactly achieve total mass $ n $ , print -1. Otherwise, print two lines. The first line should contain an integer $ d $ — the minimum number of nights needed. The next line should contain $ d $ integers, with the $ i $ -th integer representing the number of bacteria that should split on the $ i $ -th day. If there are multiple solutions, print any.

输入输出样例

输入样例 #1

3
9
11
2

输出样例 #1

3
1 0 2 
3
1 1 2
1
0

说明

In the first test case, the following process results in bacteria with total mass $ 9 $ : - Day $ 1 $ : The bacterium with mass $ 1 $ splits. There are now two bacteria with mass $ 0.5 $ each. - Night $ 1 $ : All bacteria's mass increases by one. There are now two bacteria with mass $ 1.5 $ . - Day $ 2 $ : None split. - Night $ 2 $ : There are now two bacteria with mass $ 2.5 $ . - Day $ 3 $ : Both bacteria split. There are now four bacteria with mass $ 1.25 $ . - Night $ 3 $ : There are now four bacteria with mass $ 2.25 $ . The total mass is $ 2.25+2.25+2.25+2.25=9 $ . It can be proved that $ 3 $ is the minimum number of nights needed. There are also other ways to obtain total mass 9 in 3 nights. $ $ In the second test case, the following process results in bacteria with total mass $ 11 $ : - Day $ 1 $ : The bacterium with mass $ 1 $ splits. There are now two bacteria with mass $ 0.5 $ . - Night $ 1 $ : There are now two bacteria with mass $ 1.5 $ . - Day $ 2 $ : One bacterium splits. There are now three bacteria with masses $ 0.75 $ , $ 0.75 $ , and $ 1.5 $ . - Night $ 2 $ : There are now three bacteria with masses $ 1.75 $ , $ 1.75 $ , and $ 2.5 $ . - Day $ 3 $ : The bacteria with mass $ 1.75 $ and the bacteria with mass $ 2.5 $ split. There are now five bacteria with masses $ 0.875 $ , $ 0.875 $ , $ 1.25 $ , $ 1.25 $ , and $ 1.75 $ . - Night $ 3 $ : There are now five bacteria with masses $ 1.875 $ , $ 1.875 $ , $ 2.25 $ , $ 2.25 $ , and $ 2.75 $ . The total mass is $ 1.875+1.875+2.25+2.25+2.75=11 $ . It can be proved that $ 3 $ is the minimum number of nights needed. There are also other ways to obtain total mass 11 in 3 nights. $ $ In the third test case, the bacterium does not split on day $ 1 $ , and then grows to mass $ 2 $ during night $ 1 $ .