Number of Parallelograms(求平行四边形个数)
4 seconds
256 megabytes
standard input
standard output
You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.
The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.
Each of the next n lines contains two integers (xi, yi) (0 ≤ xi, yi ≤ 109) — the coordinates of the i-th point.
Print the only integer c — the number of parallelograms with the vertices at the given points.
4 0 1 1 0 1 1 2 0
1
题解:平行四边形的中点是确定的,那么根据两两边的中点,如果有重合那么可以组成平行四边形;
代码:
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner; public class 平行四边形数 {
static Scanner cin = new Scanner(System.in);
public static void main(String[] argvs){
int N = cin.nextInt();
Point[] a = new Point[N];
for(int i = ; i < N; i++){
a[i] = new Point();
a[i].x = cin.nextDouble();
a[i].y = cin.nextDouble();
//a[i].Put();
}
Point[] b = new Point[N * (N + )/];
int tp = ;
for(int i = ; i < N*(N + )/; i++)
b[i] = new Point();
for(int i = ; i < N; i++){
for(int j = i + ; j < N; j++){
if(a[i].x == a[j].x && a[i].y == a[j].y)
continue;
b[tp++] = Point.getp(a[i], a[j]);
}
} Arrays.sort(b, , tp, new PointComparator());
// System.out.println("tp = " + tp);
int ans = , cnt = ;
for(int i = ; i < tp; i++){
// System.out.println(b[i].x + " " + b[i].y);
if(b[i].x == b[i - ].x && b[i].y == b[i - ].y){
cnt++;
}
else{
ans += cnt * (cnt - ) / ;
cnt = ;
}
}
System.out.println(ans);
}
} class Point{
static Scanner cin = new Scanner(System.in);
public double x, y;
public void Put(){
x = cin.nextInt();
y = cin.nextInt();
// System.out.println("***" + this.x + " " + this.y);
}
static Point getp(Point a, Point b){
Point c = new Point();
c.x = (a.x + b.x) / ;
c.y = (a.y + b.y) / ;
// System.out.println("***" + c.x + " " + c.y);
return c;
}
}
class PointComparator implements Comparator{
public int compare(Object x, Object y){
Point a = (Point)x;
Point b = (Point)y;
if(a.x != b.x){
if(a.x <= b.x)
return -;
else
return ;
}
else{
if(a.y <= b.y)
return -;
else
return ;
}
}
}
Number of Parallelograms(求平行四边形个数)的更多相关文章
- 【CodeForces 660D】Number of Parallelograms(n个点所能组成的最多平行四边形数量)
You are given n points on a plane. All the points are distinct and no three of them lie on the same ...
- Parallelogram Counting(平行四边形个数,思维转化)
1058 - Parallelogram Counting PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit ...
- D. Number of Parallelograms
D. Number of Parallelograms 原题链接 time limit per test 4 seconds memory limit per test 256 megabytes Y ...
- POJ 1791 Parallelogram Counting(求平行四边形数量)
Description There are n distinct points in the plane, given by their integer coordinates. Find the n ...
- Number of Parallelograms CodeForces - 660D (几何)
Number of Parallelograms CodeForces - 660D You are given n points on a plane. All the points are dis ...
- 75 int类型数组中除了一个数出现一次或两次以外,其他数都出现三次,求这个数。[2行核心代码]
[本文链接] http://www.cnblogs.com/hellogiser/p/single-number-of-array-with-other-three-times.html [题目] i ...
- js求连个数之间的数字
整理出自项目中一个需求,求两个数之间的数字. const week = function(arr,arr2){ let a=parseInt(arr); let b=parseInt(arr2); l ...
- Educational Codeforces Round 11 D. Number of Parallelograms 暴力
D. Number of Parallelograms 题目连接: http://www.codeforces.com/contest/660/problem/D Description You ar ...
- codeforces 660D D. Number of Parallelograms(计算几何)
题目链接: D. Number of Parallelograms time limit per test 4 seconds memory limit per test 256 megabytes ...
随机推荐
- iOS 部分问题总结2 - 苹果审核篇
iOS 部分问题总结(二) - 苹果审核篇 1. 记录下5.1新规后上传被拒的问题排查和解决过程. 几天前,最新一次的更新被拒了,提示Invaild Binary.好在苹果同时发来了说明邮件做了详细说 ...
- javascript自定义浏览器右键菜单
javascript自定义浏览器右键菜单 在书上看到document对象还有一个contextmenu事件,但是不知为什么w3school中找不到这个耶... 利用这个特性写了个浏览器的右键菜单, ...
- ZOJ 3430 Detect the Virus 【AC自动机+解码】
解码的那些事儿,不多说. 注意解码后的结果各种情况都有,用整数数组存储,char数组会超char类型的范围(这个事最蛋疼的啊)建立自动机的时候不能用0来判断结束. #include <cstdi ...
- 2013 长沙网络赛 B 题 Bizarre Routine
题解 http://blog.csdn.net/u010257508/article/details/11936129 #include <iostream> #include <c ...
- Homebrew新一代OS X套件管理工具 高速安装Git
在Mac上安装一些开源程序,除了自己下下载.编译( ./configure && make && make install) 之外,通常最方便的选择就是用套件管理工具來 ...
- c++11: trailing return type in functions(函数返回类型后置)
In C++03, the return type of a function template cannot be generalized if the return type relies on ...
- Sybase安装后的配置工作
1.配置数据库参数 配置sybase数据库使用的最大内存 用isql命令行实用工具登录sybase数据库服务器,其中的servername是$SYBASE/interfaces文件中配置的sybase ...
- 程序员的家!我终于拥有自己的blog了!!!
经过多次提交诚恳的家园申请,终于得到了审核通过!今天就开始了我的.net成长之路!!!
- JS--显示类型转换Number—隐式类型转换
显示类型转换 (强制类型转换):Number()parseInt()parseFloat() Number是整体转换--能够把一个看起来像数字的字符串转成数字--尽量去转换能转的都转了 var a = ...
- mysql 中的数据类型
unsigned 既为非负数,用此类型可以增加数据长度! 例如如果 tinyint最大是127,那 tinyint unsigned 最大 就可以到 127 * ...