FZU 2140 Forever 0.5（找规律，几何）

Problem 2140 Forever 0.5
Accept: 371 Submit: 1307 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

以原点为圆心，半径为1的圆内，以原点为顶点，变成为1的正三角形另外两个点在圆上，你会发现，两个点之间的那段弧，上的所有点都是满足条件的，所以只要三个顶点分别是正三角形的三个顶点，其余的点在弧上，都是正确的

#include <iostream>#include <string.h>#include <stdlib.h>#include <algorithm>#include <math.h>#include <stdio.h>using namespace std;int n;int t;int main(){    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        if(n<=3)            printf("No\n");        else        {            printf("Yes\n");            printf("0.000000 0.000000\n");            printf("0.500000 %.6f\n",-1.0*sqrt(3.0)/2);            printf("-0.500000 %.6f\n",-1.0*sqrt(3.0)/2);            for(int i=1;i<=n-3;i++)                printf("-0.000000 -1.000000\n");        }    }            return 0;}