A14642. Mathematical Addition
编程题
普及/提高-
知识点
题目描述
Ivan decided to prepare for the test on solving integer equations. He noticed that all tasks in the test have the following form:
- You are given two positive integers $u$ and $v$, find any pair of integers (**not necessarily positive**) $x$, $y$, such that:
$$\frac{x}{u} + \frac{y}{v} = \frac{x + y}{u + v}.$$
- The solution $x = 0$, $y = 0$ is forbidden, so you should find any solution with $(x, y) \neq (0, 0)$.
Please help Ivan to solve some equations of this form.
- You are given two positive integers $u$ and $v$, find any pair of integers (**not necessarily positive**) $x$, $y$, such that:
$$\frac{x}{u} + \frac{y}{v} = \frac{x + y}{u + v}.$$
- The solution $x = 0$, $y = 0$ is forbidden, so you should find any solution with $(x, y) \neq (0, 0)$.
Please help Ivan to solve some equations of this form.
输入格式
The first line contains a single integer $t$ ( $1 \leq t \leq 10^3$ ) — the number of test cases. The next lines contain descriptions of test cases.
The only line of each test case contains two integers $u$ and $v$ ( $1 \leq u, v \leq 10^9$ ) — the parameters of the equation.
The only line of each test case contains two integers $u$ and $v$ ( $1 \leq u, v \leq 10^9$ ) — the parameters of the equation.
输出格式
For each test case print two integers $x$ , $y$ — a possible solution to the equation. It should be satisfied that $-10^{18} \leq x, y \leq 10^{18}$ and $(x, y) \neq (0, 0)$ .
We can show that an answer always exists. If there are multiple possible solutions you can print any.
We can show that an answer always exists. If there are multiple possible solutions you can print any.
输入输出样例
输入 #1
4 1 1 2 3 3 5 6 9
输出 #1
-1 1 -4 9 -18 50 -4 9
说明/提示
In the first test case: $\frac{-1}{1} + \frac{1}{1} = 0 = \frac{-1 + 1}{1 + 1}$ .
In the second test case: $\frac{-4}{2} + \frac{9}{3} = 1 = \frac{-4 + 9}{2 + 3}$ .
In the third test case: $\frac{-18}{3} + \frac{50}{5} = 4 = \frac{-18 + 50}{3 + 5}$ .
In the fourth test case: $\frac{-4}{6} + \frac{9}{9} = \frac{1}{3} = \frac{-4 + 9}{6 + 9}$ .
In the second test case: $\frac{-4}{2} + \frac{9}{3} = 1 = \frac{-4 + 9}{2 + 3}$ .
In the third test case: $\frac{-18}{3} + \frac{50}{5} = 4 = \frac{-18 + 50}{3 + 5}$ .
In the fourth test case: $\frac{-4}{6} + \frac{9}{9} = \frac{1}{3} = \frac{-4 + 9}{6 + 9}$ .