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Codeforces Round #261 (Div. 2)459A. Pashmak and Garden（数学题）

2024-07-17 23:23:38   226人阅读

题目链接：http://codeforces.com/problemset/problem/459/A


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

题意：

给出两个点的坐标，问再补充两个点能否形成正方形，能则输出其余两点的坐标，否则输出-1。


PS:

昨晚手贱了（做题的姿势不太对，），本来是一道很简单的题，做得太急忘了考虑斜率为-1的情况（当时竟然过了）；最后就被别人Hack了，无语的是当时我已经锁了这题，只能眼睁睁的看着被Hack，这还是第一次被别人Hack掉。真是智商捉急啊！


被别人一个案例3 5 5 3直接打屎！




代码如下：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
const double eps = 1e-9;
#define INF 1e18
//typedef long long LL;
//typedef __int64 LL;
int main()
{
    int x1,x2,y1,y2;
    int x3,y3,x4,y4;

    while(~scanf("%d%d%d%d",&x1,&y1,&x2,&y2))
    {
        int t;
        if(x1 == x2 )
        {
            t = y1-y2;
            if(t < 0)
                t = -t;
            x3 = x1+t,x4 = x2+t;
            y3 = y1, y4 = y2;
            printf("%d %d %d %d\n",x3,y3,x4,y4);
            continue;
        }
        if(y1 == y2)
        {
            t = x1-x2;
            if(t < 0)
                t = -t;
            y3 = y1+t,y4 = y2+t;
            x3 = x1, x4 = x2;
            printf("%d %d %d %d\n",x3,y3,x4,y4);
            continue;
        }
        //if((y2-y1) == (x2-x1))//斜率为1
        if((y2-y1) == (x2-x1) ||(y2-y1) == -(x2-x1))//斜率为1或者-1
        {
            x3 = x1, y3 = y2;
            x4 = x2, y4 = y1;
            printf("%d %d %d %d\n",x3,y3,x4,y4);
            continue;
        }

        printf("-1\n");
    }
    return 0;
}




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