close
題目概要：
輸入4個座標，其中有2個座標相同，請找平形四邊形的第4個座標。
解題方向：
利用對角線相加的原理下去解即可。
Ps UVA Online Judge 測資會用到Double。
程式碼：
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
| //Java | |
| import java.util.Scanner; | |
| class uva10242{ | |
| public static void main(String args[]){ | |
| Scanner sc=new Scanner(System.in); | |
| while(sc.hasNextDouble()){ | |
| double point[][]=new double[4][2]; | |
| double same[]=new double[2]; | |
| for(int i=0;i<point.length;i++){ | |
| for(int j=0;j<point[i].length;j++){ | |
| point[i][j]=sc.nextDouble(); //儲存座標 | |
| } | |
| //找出輸入中相同的座標 | |
| for(int j=0;j<i;j++){ | |
| if(point[j][0]==point[i][0] && point[j][1]==point[i][1]){ | |
| same[0]=point[i][0]; | |
| same[1]=point[i][1]; | |
| } | |
| } | |
| } | |
| //兩對角線相加答案皆一樣的想法去解 | |
| double x=0,y=0; | |
| for(int i=0;i<point.length;i++){ | |
| if(!(point[i][0]==same[0] && point[i][1]==same[1])){ | |
| x+=point[i][0]; | |
| y+=point[i][1]; | |
| } | |
| } | |
| System.out.printf("%.3f %.3f",x-same[0],y-same[1]); | |
| System.out.println(""); | |
| } | |
| } | |
| } |
文章標籤
全站熱搜
留言列表
