zhber
有好多做过的题没写下来,如果我还能记得就补吧
随笔 - 501,  文章 - 0,  评论 - 18,  阅读 - 13万

 

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

 

大家好我是紫名选手T T第二次akdiv2了

第一题神模拟……有一个边和xy轴平行的正方形,给两个点,要输出另两个点,不存在就输出-1

然后就是各种判断啦T T

x1==x2的时候分一类,y1==y2的时候分一类,否则是对角线的位置分一类,然后判一下-1就好了

直接把x1==x2和y1==y2的输出一大堆拷过去再改的,然后忘记交换x和y了QAQ

复制代码
#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
复制代码

 

posted on   zhber  阅读(201)  评论(0编辑  收藏  举报
编辑推荐:
· 智能桌面机器人:用.NET IoT库控制舵机并多方法播放表情
· Linux glibc自带哈希表的用例及性能测试
· 深入理解 Mybatis 分库分表执行原理
· 如何打造一个高并发系统?
· .NET Core GC压缩(compact_phase)底层原理浅谈
阅读排行:
· 新年开篇:在本地部署DeepSeek大模型实现联网增强的AI应用
· DeepSeek火爆全网,官网宕机?本地部署一个随便玩「LLM探索」
· Janus Pro:DeepSeek 开源革新,多模态 AI 的未来
· 上周热点回顾(1.20-1.26)
· 【译】.NET 升级助手现在支持升级到集中式包管理

< 2025年1月 >
29 30 31 1 2 3 4
5 6 7 8 9 10 11
12 13 14 15 16 17 18
19 20 21 22 23 24 25
26 27 28 29 30 31 1
2 3 4 5 6 7 8
点击右上角即可分享
微信分享提示