首頁 > web前端 > html教學 > Codeforces Round #261(Div.2) A. Pashmak and Garden_html/css_WEB-ITnose

Codeforces Round #261(Div.2) A. Pashmak and Garden_html/css_WEB-ITnose

WBOY
發布: 2016-06-24 11:59:34
原創
1003 人瀏覽過

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
登入後複製

output

1 0 1 1
登入後複製

input

0 0 1 1
登入後複製

output

0 1 1 0
登入後複製

input

0 0 1 2
登入後複製

output

-1
登入後複製
#include<stdio.h>#include<math.h>int main(){    int x1,x2,y1,y2,x3,x4,y3,y4,h1,h2;    while(scanf("%d%d%d%d",&x1,&y1,&x2,&y2)!=EOF)    {        h1=x1-x2;h2=y1-y2;        h1=fabs(h1);        h2=fabs(h2);        if(h1!=0&&h2!=0&&h1!=h2)        {             printf("-1\n");             continue;        }        if(h1==0)        {            y3=y1;y4=y2;            x3=x1+h2;            x4=x2+h2;        }        if(h2==0)        {            x3=x1;x4=x2;            y3=y1+h1;            y4=y2+h1;        }        if(h1==h2)        {            x3=x1;x4=x2;            if(y1>y2)            {                y3=y1-h1;                y4=y2+h1;            }            else            {                y3=y1+h1;                y4=y2-h1;            }        }        printf("%d %d %d %d\n",x3,y3,x4,y4);    }    return 0;}</math.h></stdio.h>
登入後複製

相關標籤:
來源:php.cn
本網站聲明
本文內容由網友自願投稿,版權歸原作者所有。本站不承擔相應的法律責任。如發現涉嫌抄襲或侵權的內容,請聯絡admin@php.cn
熱門教學
更多>
最新下載
更多>
網站特效
網站源碼
網站素材
前端模板