cf459A Pashmak and Garden

A. Pashmak and Garden time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Pashmak has fallen in love with an attractive girl called Parmida since one year ago...

Today, Pashmak set up a meeting with his partner in a romantic garden. Unfortunately, Pashmak has forgotten where the garden is. But he remembers that the garden looks like a square with sides parallel to the coordinate axes. He also remembers that there is exactly one tree on each vertex of the square. Now, Pashmak knows the position of only two of the trees. Help him to find the position of two remaining ones.

Input

The first line contains four space-separated x1, y1, x2, y2 ( - 100 ≤ x1, y1, x2, y2 ≤ 100) integers, where x1 and y1 are coordinates of the first tree and x2 and y2 are coordinates of the second tree. It's guaranteed that the given points are distinct.

Output

If there is no solution to the problem, print -1. Otherwise print four space-separated integers x3, y3, x4, y4 that correspond to the coordinates of the two other trees. If there are several solutions you can output any of them.

Note that x3, y3, x4, y4 must be in the range ( - 1000 ≤ x3, y3, x4, y4 ≤ 1000).

Sample test(s) input

0 0 0 1
      

output

1 0 1 1
      

input

0 0 1 1
      

output

0 1 1 0
      

input

0 0 1 2
      

output

-1      

大家好我是紫名選手T T第二次akdiv2了

第一題神模拟……有一個邊和xy軸平行的正方形，給兩個點，要輸出另兩個點，不存在就輸出-1

然後就是各種判斷啦T T

x1==x2的時候分一類，y1==y2的時候分一類，否則是對角線的位置分一類，然後判一下-1就好了

直接把x1==x2和y1==y2的輸出一大堆拷過去再改的，然後忘記交換x和y了QAQ

#include<iostream>
#include<cstdio>
#define LL long long
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline int abs(int a)
{return a<0?-a:a;}
int x1,x2,y1,y2;
int main()
{
    x1=read();y1=read();x2=read();y2=read();
    if (x1==x2)
    {
        printf("%d %d %d %d",abs(y2-y1)+x1,y1,abs(y2-y1)+x1,y2);
    }else
    if (y1==y2)
    {
        printf("%d %d %d %d",x1,abs(x2-x1)+y1,x2,abs(x2-x1)+y1);
    }else 
    {
        if (abs(x1-x2)!=abs(y1-y2)){printf("-1");return 0;}
        printf("%d %d %d %d",x1,y2,x2,y1);
    }
}      

cf459A

轉載于:https://www.cnblogs.com/zhber/p/4035920.html