Minimal Square

文章目录

一、A. Minimal Square

总结

一、A. Minimal Square

本题链接：A. Minimal Square

题目：

A. Minimal Square

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

Find the minimum area of a square land on which you can place two identical rectangular a×b houses. The sides of the houses should be parallel to the sides of the desired square land.

Formally,

You are given two identical rectangles with side lengths a and b (1≤a,b≤100) — positive integers (you are given just the sizes, but not their positions).

Find the square of the minimum area that contains both given rectangles. Rectangles can be rotated (both or just one), moved, but the sides of the rectangles should be parallel to the sides of the desired square.

Two rectangles can touch each other (side or corner), but cannot intersect. Rectangles can also touch the sides of the square but must be completely inside it. You can rotate the rectangles. Take a look at the examples for a better understanding.

Input

The first line contains an integer t (1≤t≤10000) —the number of test cases in the input. Then t test cases follow.

Each test case is a line containing two integers a, b (1≤a,b≤100) — side lengths of the rectangles.

Output

Print t answers to the test cases. Each answer must be a single integer — minimal area of square land, that contains two rectangles with dimensions a×b.

Example

input

8

3 2

4 2

1 1

3 1

4 7

1 3

7 4

100 100

output

16

16

4

9

64

9

64

40000

Note

Below are the answers for the first two test cases:

本博客给出本题截图：

题意：t组数据，每次数据给一个长方形的长和宽，现在把两个 相同 的长方形放入到一个正方形内部，可以旋转长方形但旋转必须满足长方形的边和正方形的边相平行，两个正方形不可以重叠，求正方形面积最小值

AC代码

#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    int t;
    cin >> t;
    while (t -- )
    {
        int l, r;
        cin >> l >> r;
        if (l < r)
            swap(l, r);
        int len = max(r * 2, l);
        cout << len * len << endl;
    }
    return 0;
}

总结

水题，不解释