Number of Triplets

A. Number of Triplets
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given n points on a plane. All points are different.

Find the number of different groups of three points (A, B, C) such that point B is the middle of segment AC.

The groups of three points are considered unordered, that is, if point B is the middle of segment AC, then groups (A, B, C) and (C, B, A) are considered the same.

Input
The first line contains a single integer n (3 ≤ n ≤ 3000) — the number of points.

Next n lines contain the points. The i-th line contains coordinates of the i-th point: two space-separated integers xi, yi ( - 1000 ≤ xi, yi ≤ 1000).

It is guaranteed that all given points are different.

Output
Print the single number — the answer to the problem.

Examples
inputCopy
3
1 1
2 2
3 3
outputCopy
1
inputCopy
3
0 0
-1 0
0 1
outputCopy
0

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iomanip>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;struct node{int x,y;friend bool operator < (node a,node b){if(a.x!=b.x)return a.x<b.x;return a.y<b.y;}friend bool operator ==(node a,node b){return a.x==b.x&&a.y==b.y;}
}p[3001];
int n,ans;void judge(int l,int r){while(l<=r){int mid=(l+r)>>1;p[mid]<p[0]?l=mid+1:r=mid-1;}if(p[l]==p[0]||p[r]==p[0])ans++;
}int main(){while(cin>>n){for(int i=1;i<=n;i++)cin>>p[i].x>>p[i].y;sort(p+1,p+1+n);ans=0;for(int i=1;i<n;i++){for(int j=i+1;j<=n;j++){p[0].x=(p[i].x+p[j].x)/2;p[0].y=(p[i].y+p[j].y)/2;if(p[0].x*2==p[i].x+p[j].x&&p[0].y*2==p[i].y+p[j].y){judge(1,n);}}}cout<<ans<<endl;}return 0;
}

Number of Triplets相关推荐

[Swift]LeetCode611. 有效三角形的个数 | Valid Triangle Number
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ ➤微信公众号:山青咏芝(shanqingyongzhi) ➤博客园地址:山青咏芝(https://www.cnblog ...
语义分割--Mix-and-Match Tuning for Self-Supervised Semantic Segmentation
Mix-and-Match Tuning for Self-Supervised Semantic Segmentation AAAI Conference on Artificial Intelli ...
Java使用RSA加密解密签名及校验
RSA加密解密类: import java.io.BufferedReader; import java.io.BufferedWriter; import java.io.FileReader; i ...
linux下部署tomcat没起成功,linux下部署tomcat 上线jpress博客系统
tomcat Tomcat服务器是一个免费的开放源代码的Web应用服务器,属于轻量级应用服务器. tomcat有三个端口开始部署安装tomcat 第一步:下载tomcat 安装包 tomcat官网 ...
C++ 使用OpenSSL 基于SHA1摘要的RSA签名及验签与Java平台互通
文章目录准备 C++ Java RSASignature.java RSAEncrypt.java Base64.java 准备配置OpenSSL环境配置VS2015环境生成公私秘钥然后你们 ...
Codeforces Beta Round #57 (Div. 2)E---Enemy is weak(树状数组+离散化)
The Romans have attacked again. This time they are much more than the Persians but Shapur is ready t ...
Android支付接入（一）：支付宝
转载之前我想深深地感谢屌丝哥相信相同过App获取利润的都会需要接入计费SDK,下边就跟大家走一遍完整的支付宝SDK接入,支付系列均通过计费Button触发,计费所有代码均放到一个java文件实现,这 ...
UPC-2021个人训练赛第20场-部分题解
目录 RGB Triplets 题目描述输入输出样例输入 Copy 样例输出 Copy 提示 Select Half 题目描述输入输出样例输入 Copy 样例输出 Copy 提示心灵的抚 ...
java调用爱贝支付
目录 1.爱贝支付流程 2.接口说明 2.1下单-获取transid 2.2返回参数及格式 3.代码实现 3.1项目结构 3.2OrderMain 3.3OrderTest 3.4RSA 3.5Bas ...
RSA非对称加密（以Android和java为例）
一.概念 1.1目的由于工作中用到了趟过了许多的坑所以深有体会,所以想写个笔记怕以后忘记了,好到时候回来看看. 首先是一定要知道最基础的概念,一定要知道最基础的概念,一定要知道最基础的概念,以免被误 ...

Number of Triplets

Number of Triplets相关推荐

最新文章

热门文章