FZU 2140 Forever 0.5

 Problem 2140 Forever 0.5

Accept: 36    Submit: 113    Special Judge
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 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

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 ：给你一个数n，让你找出n个点，满足一下关系：

• 任意两点的距离不大于1.0
• 所有点到原点的距离不大于1.0
• 恰好有N对点的距离为1.0
• 由这些点构成的n边形的面积不小于0.5
• 由这些点构成的n边形的面积不大于0.75

如果有就输出yes加上这n个点，如果没有就输出no

思路 ： 这个题一开始看样例觉得好复杂，其实画个图推一下倒是可以看出来，要满足上边的条件至少要是4个点，3个点的话是一个等边三角形，面积不符合。因为条件中老是提到1，其实就是一个半径为1的圆以原点为圆心。然后以原点和x轴画一个边长为1的等边三角形，这样的话就有三个点了，其实前四个点都是可以确定的，然后剩下的点从圆上找就可以了，主要是别离那三个点的距离大于1即可，因为圆上的点到圆心的距离都为1，其实就是将圆离散化。

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#include <iostream>

#include <algorithm>

#include <math.h>

using namespace std ;

double x[105],y[105] ;

const double temp = 0.005 ;

void chart()

{

    x[0] = 0,y[0] = 0 ;

    x[1] = 1,y[1] = 0 ;

    x[2] = 0.5,y[2] = sqrt(1.0-0.25) ;

    x[3] = 0.5 ,y[3] = y[2]-1 ;

    for(int i = 4 ; i < 105 ; i++)

    {

        y[i] = i*temp ;

        x[i] = sqrt(1-y[i]*y[i]) ;

    }

}

int main()

{

    int T,n ;

    chart() ;

    scanf("%d",&T) ;

    while(T--)

    {

        scanf("%d",&n) ;

        if(n < 4)

            printf("No\n") ;

        else

        {

            printf("Yes\n") ;

            for(int i = 0 ; i < n ; i++)

                printf("%.6lf %.6lf\n",y[i],x[i]) ;

        }

    }

    return 0 ;

}

View Code