FZU 2140 Forever 0.5 (几何构造)

Forever 0.5

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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

3

2

3

5

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.

题意：找n个点满足上述五个条件。

题解：n=1和n=2的时候不能构成多边形输出No，n=3的时候最大是边长为1的等边三角形，面积为4分之根号3小于0.5不符合条件输出No。当n=4的时候，在等边三角形的一个顶点的高的方向加一个点，距离该顶点的距离为1即可满足五个条件，如图所示。

AC=AB=BC=BD=1。多边形外侧是一个以(0,0)为圆心半径为1的圆。

当n>=5时，在BC弧上加点即可，Smax=S扇形ABC+S三角形ACD=pi/6+1*(1-sqrt(3)/2)/2=0.59058。符合条件。

角bac是60度也就是pi/3，新加入的点的弧度f为pi/3-0.01*(n-4)即可，n最大100，最多加入96个点不会溢出。

新加入点的横坐标就是cos(f)，纵坐标就是sin(f)。用y=sqrt(1-x*x)算也行。

#include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;
#define pi 3.1415926535897931
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        if(n<=)
        {
            printf("No\n");
        }
        else
        {
            printf("Yes\n"); //开始的时候忘了输出Yes错了三遍
            double a=sqrt()/2.0;
            printf("0 0\n");
            printf("1 0\n");
            printf("0.5 %.6lf\n",a);
            printf("0.5 %.6lf\n",a-);
            double b=pi/3.0;
            for(int i=;i<n-;i++)
            {
                b=b-0.01;
                printf("%.6lf %.6lf\n",cos(b),sin(b));
            }
        }
    }
    return ;
}