2020牛客暑期多校训练营（第四场）F

题目描述

ZYB has a so-called smart brain. He can always point out the key-point in a complex problem.
There are two parallel lines $AB$ and $CD$ in a plane. $A,B,C,D$ are all distinct points.
You only know the Euclidean Distances between $AC,AD,BC,BD$ . But you don't know the exact order of points (i.e. You don't know whether it's $AB$ $// CD$ or $AB// DC$ ).
Could you determine the order of points quickly, like the ZYB does?

输入描述

The input contains multiple cases. The first line of the input contains a single integer $T$ $(1 \leqslant T \leqslant 100)$ , the number of cases.
For each case, there are four integers $a,b,c,d$ $(1 \leqslant a,b,c,d \leqslant 1000)$ in a line, indicating the distances between $AC,AD,BC,BD$ .
It is guaranteed that each case corresponds to a valid solution.

输出描述

For each case, output AB//CD (Quotation marks) if $AB//CD$ , or output AB//DC (Quotation marks) if $AB //DC$ .

示例1

输入

2  
3 5 5 3
5 3 3 5

输出

AB//CD
AB//DC

分析

由三角形三边关系， $p+s>x$ ， $r+q>y$ ，故 $p+q+r+s>x+y$ 。

当 $AB//CD$ ，有 $AD+BC>AC+BD$ 。

当 $AB//DC$ ，有 $AC+BD>AD+BC$ 。

找到了完全相反的两个结论，可以作为判断的依据。

代码

/******************************************************************
Copyright: 11D_Beyonder All Rights Reserved
Author: 11D_Beyonder
Problem ID: 2020牛客暑期多校训练营（第四场） Problem F
Date: 8/15/2020
Description: Computational Geometry
*******************************************************************/
#include<iostream>
#include<cstdio>
using namespace std;
int main(){
    int _;
    for(cin>>_;_;_--){
        int AC,AD,BC,BD;
        scanf("%d%d%d%d",&AC,&AD,&BC,&BD);
        puts(AD+BC>AC+BD?"AB//CD":"AB//DC");
    }
    return 0;
}