2015ACM/ICPC亚洲区长春站 G hdu 5533 Dancing Stars on Me
Dancing Stars on Me
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 186 Accepted Submission(s): 124
Formally, a regular polygon is a convex polygon whose angles are all equal and all its sides have the same length. The area of a regular polygon must be nonzero. We say the stars can form a regular polygon if they are exactly the vertices of some regular polygon. To simplify the problem, we project the sky to a two-dimensional plane here, and you just need to check whether the stars can form a regular polygon in this plane.
1≤T≤300
3≤n≤100
−10000≤xi,yi≤10000
All coordinates are distinct.
3
0 0
1 1
1 0
4
0 0
0 1
1 0
1 1
5
0 0
0 1
0 2
2 2
2 0
YES
NO
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <iostream>
#include <map>
#include <set>
#include <algorithm>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
using namespace std;
typedef long long LL;
typedef double DB;
#define MIT (2147483647)
#define MLL (1000000000000000001LL)
#define INF (1000000001)
#define For(i, s, t) for(int i = (s); i <= (t); i ++)
#define Ford(i, s, t) for(int i = (s); i >= (t); i --)
#define Rep(i, n) for(int i = (0); i < (n); i ++)
#define Repn(i, n) for(int i = (n)-1; i >= (0); i --)
#define mk make_pair
#define ft first
#define sd second
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define sz(x) ((int) (x).size())
inline void SetIO(string Name)
{
string Input = Name + ".in";
string Output = Name + ".out";
freopen(Input.c_str(), "r", stdin);
freopen(Output.c_str(), "w", stdout);
} inline int Getint()
{
char ch = ' ';
int Ret = ;
bool Flag = ;
while(!(ch >= '' && ch <= ''))
{
if(ch == '-') Flag ^= ;
ch = getchar();
}
while(ch >= '' && ch <= '')
{
Ret = Ret * + ch - '';
ch = getchar();
}
return Flag ? -Ret : Ret;
} const int N = ;
struct Point
{
int x, y;
} Arr[N];
int n; inline void Solve(); inline void Input()
{
int TestNumber = Getint();
while(TestNumber--)
{
n = Getint();
For(i, , n)
{
Arr[i].x = Getint();
Arr[i].y = Getint();
}
Solve();
}
} inline LL Sqr(int x) {
return 1LL * x * x;
} inline int Multi(const Point &O, const Point &A, const Point &B)
{
int X1 = A.x - O.x, X2 = B.x - O.x, Y1 = A.y - O.y, Y2 = B.y - O.y;
return X1 * Y2 - X2 * Y1;
} inline LL GetDist(const Point &A, const Point &B)
{
return Sqr(B.x - A.x) + Sqr(B.y - A.y);
} inline bool Compare(const Point &A, const Point &B)
{
int Det = Multi(Arr[], A, B);
if(Det) return Det > ;
LL Dist1 = GetDist(Arr[], A), Dist2 = GetDist(Arr[], B);
return Dist1 < Dist2;
} inline void Solve()
{
For(i, , n)
if(Arr[i].x < Arr[].x || (Arr[i].x == Arr[].x && Arr[i].y < Arr[].y))
swap(Arr[i], Arr[]);
sort(Arr + , Arr + + n, Compare); Arr[n + ] = Arr[], Arr[n + ] = Arr[];
bool Flag = ;
int Tmp, Dist;
For(i, , n)
{
int Det = Multi(Arr[i], Arr[i + ], Arr[i + ]);
LL Dist1 = GetDist(Arr[i], Arr[i + ]);
LL Dist2 = GetDist(Arr[i + ], Arr[i + ]);
if(Det <= || Dist1 != Dist2)
{
puts("NO");
return ;
}
if(Flag)
{
if(Tmp != Det || Dist1 != Dist)
{
puts("NO");
return ;
}
}
else Flag = , Tmp = Det, Dist = Dist1;
}
puts("YES");
} int main()
{
Input();
//Solve();
return ;
}
2015ACM/ICPC亚洲区长春站 G hdu 5533 Dancing Stars on Me的更多相关文章
- 2015ACM/ICPC亚洲区长春站 F hdu 5533 Almost Sorted Array
Almost Sorted Array Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Ot ...
- 2015ACM/ICPC亚洲区长春站 B hdu 5528 Count a * b
Count a * b Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Tot ...
- 2015ACM/ICPC亚洲区长春站 L hdu 5538 House Building
House Building Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) ...
- 2015ACM/ICPC亚洲区长春站 J hdu 5536 Chip Factory
Chip Factory Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)T ...
- 2015ACM/ICPC亚洲区长春站 H hdu 5534 Partial Tree
Partial Tree Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)To ...
- 2015ACM/ICPC亚洲区长春站 E hdu 5531 Rebuild
Rebuild Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total S ...
- 2015ACM/ICPC亚洲区长春站 A hdu 5527 Too Rich
Too Rich Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total ...
- hdu 5533 Dancing Stars on Me(数学,水)
Problem Description The sky was brushed clean by the wind and the stars were cold in a black sky. Wh ...
- hdu 5533 Dancing Stars on Me
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5533 Dancing Stars on Me Time Limit: 2000/1000 MS (Ja ...
随机推荐
- VS2010 使用时选择代码或双击时出错,点击窗口按钮后VS自动重启问题
VS2010 使用时选择代码或双击时出错崩溃,点击窗口按钮后VS自动重启问题 下载补丁,打上补丁之后,重启电脑,解决了问题. WindowsXP的下载地址:Windows XP 更新程序 (KB971 ...
- 手把手教你用Python爬虫煎蛋妹纸海量图片
我们的目标是用爬虫来干一件略污事情 最近听说煎蛋上有好多可爱的妹子,而且爬虫从妹子图抓起练手最好,毕竟动力大嘛.而且现在网络上的妹子很黄很暴力,一下接受太多容易营养不量,但是本着有人身体就比较好的套路 ...
- Resumable uploads over HTTP. Protocol specification
Valery Kholodkov <valery@grid.net.ru>, 2010 1. Introduction This document describes applicatio ...
- JQuery $(function(){})和$(document).ready(function(){})
document.ready和onload的区别——JavaScript文档加载完成事件页面加载完成有两种事件一是ready,表示文档结构已经加载完成(不包含图片等非文字媒体文件)二是onload,指 ...
- MVC 修饰标签
MVC中的修饰标签有很多用途.它以修饰标签形式应用在控制器或控制器中的动作上. 最先想到的就是AcceptVerbs标签,在创建的时候,如果导航到创建视图,但不创建,则: public ActionR ...
- python操作Excel读写--使用xlrd
一.安装xlrd模块 到python官网下载http://pypi.python.org/pypi/xlrd模块安装,前提是已经安装了python 环境. 二.使用介绍 1.导入模块 import x ...
- Android 中的Force Close
今天写程序时遇到一个问题,领导希望在点击了setting里的force close 后,程序依然能够响应以前用alarmManager注册的receiver. 在网上看到了一些文章,写的是如何建立一个 ...
- BestCoder10 1002 Revenge of GCD(hdu 5019) 解题报告
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5019 题目意思:给出 X 和 Y,求出 第 K 个 X 和 Y 的最大公约数. 例如8 16,它们的公 ...
- ubuntu 13.10 amd64安装ia32-libs
很多软件只有32位的,有的依赖32位库还挺严重的:从ubuntu 13.10已经废弃了ia32-libs,但可以使用多架构,安装软件或包apt-get install program:i386.有的还 ...
- Java性能优化权威指南-读书笔记(五)-JVM性能调优-吞吐量
吞吐量是指,应用程序的TPS: 每秒多少次事务,QPS: 每秒多少次查询等性能指标. 吞吐量调优就是减少垃圾收集器消耗的CPU周期数,从而将更多的CPU周期用于执行应用程序. CMS吞吐调优 CMS包 ...