FZU 2140 Forever 0.5

最新推荐文章于 2018-04-14 19:57:41 发布

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最新推荐文章于 2018-04-14 19:57:41 发布

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本文链接：https://blog.youkuaiyun.com/dark_master/article/details/72596372

本文探讨了如何判断平面上是否存在N个特殊点，这些点需满足特定的距离和位置约束，包括点间距离限制、与原点的距离限制、特定数量的距离为1的点对以及构成的凸包面积范围。

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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

No
No
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.

题目大意：给定一个整数N，你的任务是判断飞机上是否存在N点，以满足以下条件：
1.任何两点之间的距离不大于1.0。
2.任意点与原点（0,0）之间的距离不大于1.0。
正好有N对，他们的距离正好是1.0。
由这N点构成的凸包的面积不小于0.5。
由这N点构成的凸包的面积不大于0.75。

构造满足条件的多边形。以（0，0）为一个顶点构造一个等边三角形，然后再找出圆上与另外三个点距离等于1的点即可。

题目实例是假的= =~假的假的。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        if(n<=3) printf("No\n");
        else
        {
            printf("Yes\n");
            printf("%.6f %.6f\n",0.0,0.0);
            printf("%.6f %.6f\n",-0.5,sqrt(3.0)/2);
            printf("%.6f %.6f\n",0.5,sqrt(3.0)/2);
            int i;
            for(i=4;i<=n;i++)
            {
                printf("%.6f %.6f\n",0.0,1.0);
            }
        }
    }
    return 0;
}