FZU 2140 Forever 0.5

最新推荐文章于 2019-04-13 22:52:01 发布

原创最新推荐文章于 2019-04-13 22:52:01 发布 · 604 阅读

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本文深入探讨了人工智能领域的核心技术，包括机器学习、深度学习、自然语言处理等，并结合实际应用案例进行了详细分析。

Problem 2140 Forever 0.5

Problem Description

Given an integer N, your task is to judge whether there exist N points in the plane such that satisfy the following conditions:

1. The distance between any two points is no greater than 1.0.

2. The distance between any point and the origin (0,0) is no greater than 1.0.

3. There are exactly N pairs of the points that their distance is exactly 1.0.

4. The area of the convex hull constituted by these N points is no less than 0.5.

5. The area of the convex hull constituted by these N points is no greater than 0.75.

Input

The first line of the date is an integer T, which is the number of the text cases.

Then T cases follow, each contains an integer N described above.

1 <= T <= 100, 1 <= N <= 100

Output

For each case, output “Yes” if this kind of set of points exists, then output N lines described these N points with its coordinate.

Make true that each coordinate of your output should be a real number with AT MOST 6 digits after decimal point.

Your answer will be accepted if your absolute error for each number is no more than 10-4.

Otherwise just output “No”.

See the sample input and output for more details.

Sample Input

Sample Output

Yes

0.000000 0.525731

-0.500000 0.162460

-0.309017 -0.425325

0.309017 -0.425325

0.500000 0.162460

Hint

This problem is special judge.

这个是省赛预选赛第一题。。。

第一次做sj的题目

被样例误导憨憨憨地算起来了正多边形每个顶点的位置最后发现除了5边形其他的面积都大于0.75。。。

比赛完后查了一下题解才知道因为sj嘛随便搞搞只要符合条件就可以

然后以原点O为圆心以1为半径的圆上的点到原点距离都是1

所以在上面找就可以了

首先三角形要求是等边三角形发现面积小于0.5

然后四边形为了方便就这样取吧坐标比较好表示

然后在(0.5,sqrt(1-0.25))和(1,0)之间的弧上任取点到(0,0)的距离都是1 而到其他点的距离都小于1满足题目要求

所以昨天憨狗了

keep it simple & stupid

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
double x[111],y[111];
int T,m;
int main()
{
	x[1]=0;y[1]=0;
	x[2]=1;y[2]=0;
	x[3]=0.5;y[3]=(sqrt(1-0.5*0.5));
	x[4]=0.5;y[4]=y[3]-1;
	for(int i=5;i<=100;++i)
	{
		x[i]=0.5+0.004*i;
		y[i]=sqrt(1-x[i]*x[i]);
	}
	cin>>T;
	while(T)
	{
		T--;
		cin>>m;
		if(m<4)cout<<"No"<<endl;
		else
		{
			cout<<"Yes"<<endl;
			for(int j=1;j<=m;++j)
			printf("%.6lf %.6lf\n",x[j],y[j]);
		}
	}
	return 0;
}