Accept: 269 Submit: 934 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
Sample Output
开始半天没有读懂题意，以为这道题挺复杂的，后来读懂了一点，原来也是比较水，有一点技巧。
题意：给你n个点，这n个点是否满足下面的条件。
1.两点之间的距离不大于1
2.任意点到原点的距离不大于1
3.有n对点的距离刚好等于1
4.n个点构成的n边形的面积不小于0.5
5.n个点构成的n边形的面积不大于0.75
如果满足关系就输出yes + 这n个点，没有就输出 no。
思路：画个图可以进行推敲一下，前四个点基本上都是确定的。3个点就是一个等边三角形，（原点，x=1，y=0）但是面积不符合，所以满足条件至少需要4个点，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，然后剩下的点从圆上找就可以了，主要是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 <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
using namespace std;
double x[110],y[110];
const double temp=0.001;
void solve()
{
x[0]=0,y[0]=0;//原点
x[1]=1,y[1]=0;//第二个点
x[2]=0.5,y[2]=sqrt(1-0.25);//等边三角形的三个点
x[3]=0.5,y[3]=y[2]-1;
for(int i=4;i<110;i++)
{
y[i]=i*temp;//离散化
x[i]=sqrt(1-y[i]*y[i]);
}
}
int main()
{
int t,n;
solve();
cin>>t;
while(t--)
{
cin>>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;
}