Expression Evaluation Error

题意翻译

Bob 在黑板上写下了 $n$ 个十进制正整数，已知它们在十进制下的和为 $s$。Alice 求出了这些数在 $11$ 进制下的和。 请求出任意一组使得 Alice 求得的 $11$ 进制下的和最大的 $n$ 个数。 数据范围： - $t$ 组数据，$1\leqslant t\leqslant 100$。 - $1\leqslant s\leqslant 10^9$，$1\leqslant n\leqslant \min\{s,100\}$。 Translated by Eason_AC 2021.9.6

题目描述

On the board, Bob wrote $ n $ positive integers in [base](https://en.wikipedia.org/wiki/Positional_notation#Base_of_the_numeral_system) $ 10 $ with sum $ s $ (i. e. in decimal numeral system). Alice sees the board, but accidentally interprets the numbers on the board as base- $ 11 $ integers and adds them up (in base $ 11 $ ). What numbers should Bob write on the board, so Alice's sum is as large as possible?

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 \leq t \leq 100 $ ) — the number of test cases. The description of the test cases follows. The only line of each test case contains two integers $ s $ and $ n $ ( $ 1 \leq s \leq 10^9 $ ; $ 1 \leq n \leq \min(100, s) $ ) — the sum and amount of numbers on the board, respectively. Numbers $ s $ and $ n $ are given in decimal notation (base $ 10 $ ).

输出格式

For each test case, output $ n $ positive integers — the numbers Bob should write on the board, so Alice's sum is as large as possible. If there are multiple answers, print any of them.

输入输出样例

输入样例 #1

6
97 2
17 1
111 4
100 2
10 9
999999 3

输出样例 #1

70 27 
17 
3 4 100 4
10 90
1 1 2 1 1 1 1 1 1 
999900 90 9

说明

In the first test case, $ 70_{10} + 27_{10} = 97_{10} $ , and Alice's sum is $ $70_{11} + 27_{11} = 97_{11} = 9 \cdot 11 + 7 = 106_{10}. $ $ (Here $ x\_b $ represents the number $ x $ in base $ b $ .) It can be shown that it is impossible for Alice to get a larger sum than $ 106\_{10} $ .</p><p>In the second test case, Bob can only write a single number on the board, so he must write $ 17 $ .</p><p>In the third test case, $ 3\_{10} + 4\_{10} + 100\_{10} + 4\_{10} = 111\_{10} $ , and Alice's sum is $ $ 3_{11} + 4_{11} + 100_{11} + 4_{11} = 110_{11} = 1 \cdot 11^2 + 1 \cdot 11 = 132_{10}. $ $ It can be shown that it is impossible for Alice to get a larger sum than $ 132\_{10}$.