CodeForces - 1584A Mathematical Addition数学计算

Posted 2022-06-23 海岛Blog

A. Mathematical Addition
time limit per test 1 second
memory limit per test 256 megabytes

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)≠(0,0).
Please help Ivan to solve some equations of this form.

Input
The first line contains a single integer t (1≤t≤10³) — 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≤u,v≤10⁹) — the parameters of the equation.

Output
For each test case print two integers x, y — a possible solution to the equation. It should be satisfied that −10¹⁸≤x,y≤10¹⁸ and (x,y)≠(0,0).

We can show that an answer always exists. If there are multiple possible solutions you can print any.

Example
input
4
1 1
2 3
3 5
6 9
output
-1 1
-4 9
-18 50
-4 9

Note
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: $\frac{-4}{6}+\frac{9}{9}=\frac{1}{3}=\frac{-4+9}{6+9}.$

问题链接：CodeForces - 1584A Mathematical Addition
问题简述：（略）
问题分析：（略）

AC的C语言程序如下：

/* CodeForces - 1584A Mathematical Addition */

#include <stdio.h>

int main()

    int t;
    scanf("%d", &t);
    while (t--) 
        long long u, v;
        scanf("%lld%lld", &u, &v);
        printf("%lld %lld\\n", - u * u, v * v);
    

    return 0;