A14533. Expression Evaluation Error
编程题
普及/提高-
知识点
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?
What numbers should Bob write on the board, so Alice's sum is as large as possible?
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 $10$ ).
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 #1
6 97 2 17 1 111 4 100 2 10 9 999999 3
输出 #1
70 27 17 3 4 100 4 10 90 1 1 2 1 1 1 1 1 1 999900 90 9
说明/提示
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}$$$.