2020牛客暑期多校训练营（第四场）F题

链接：https://ac.nowcoder.com/acm/contest/5669/F
来源：牛客网

题目描述

ZYB has a so-called smart brain: He can always point out the keypoint in a complex problem.
There are two parallel lines AB and CD in a plane. A,B,C,D{A,B,C,D}A,B,C,D are all distinct points. You only know the Euclidean Distances between AC,AD,BC,BD{AC,AD,BC,BD}AC,AD,BC,BD. but you don’t know the exact order of points. (i.e. You don’t know whether it’s AB∥CDAB \parallel CDAB∥CD or AB∥DCAB \parallel DCAB∥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≤T≤100)T\ (1 \le T \le 100)T (1≤T≤100), the number of cases.
For each case, there are four integers a,b,c,d(1≤a,b,c,d≤1000)a,b,c,d(1 \le a,b,c,d \le 1000)a,b,c,d(1≤a,b,c,d≤1000) in a line, indicating the distances between AC,AD,BC,BD{AC,AD,BC,BD}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∥CDAB \parallel CDAB∥CD, or output ‘AB//DC’ (Quotation marks) if AB∥DCAB \parallel DCAB∥DC.

示例1

输入

2
3 5 5 3
5 3 3 5

2
3 5 5 3
5 3 3 5

输出

AB//CD
AB//DC

AB//CD
AB//DC

#pragma warning (disable:4996)
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <list>
#define inf 0X3f3f3f3f
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

int main()
{
 int t;
 scanf("%d", &t);
 while (t--)
 {
  int ac, ad, bc, bd;
  scanf("%d%d%d%d", &ac, &ad, &bc, &bd);
  bool ans;//1:CD, 0:DC
  if (ac < bc)//c左
  {
   if (ad < bd)//d左
   {
    if (bc > bd)
     ans = 1;
    else
     ans = 0;
   }
   else//d右
    ans = 1;
  }
  else//c右
  {
   if (ad < bd)//d左
    ans = 0;
   else//d右
   {
    if (ac < ad)
     ans = 1;
    else
     ans = 0;
   }
  }
  if (ans)
   printf("AB//CD\n");
  else
   printf("AB//DC\n");
 }
}