【构造】 FZU 2140 Forever 0.5

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题意：输入N，表示有N个点，

要求：

1.任意两点的距离≤ 1.0

2.每个点与原点的距离≤1.0

3.有N对点间的距离=1.0

4.N个点形成的面积≥0.5 ≤0.75

可以先取一个以原点为顶点，另外两个点在单位圆上的正三角形

其他的点就在圆弧BC上

将BC等分

而且N==4时 第四个点必须要在BC中点上

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <cctype>
#include <cmath>
#include <string>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
typedef long long LL;
typedef long double LD;
const double eps=1e-8;
#define pi acos(-1.0)
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
typedef pair<int, int> PI;
typedef pair<int, PI> PP;
#ifdef _WIN32
#define LLD "%I64d"
#else
#define LLD "%lld"
#endif
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}
//inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;}
//inline void print(LL x){printf(LLD, x);puts("");}
//inline void read(double &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}}
//inline void sc(LL &x){scanf(LLD, &x);}
int main()
{
    int n,t;
    cin>>t;
    while(t--)
    {
        cin>>n;
        if(n<4)
            printf("No\n");
        else
        {
            printf("Yes\n");
            printf("%.6lf %.6lf\n",0.0,0.0);
            printf("%.6lf %.6lf\n",1.0,0.0);
            printf("%.6lf %.6lf\n",0.5,sqrt(3.0)/2.0);
            double st;
            for(int i=1;i<=n-3;i++)
            {
                st=(pi/3.0/(n-2))*i;
                double xx=cos(st);
                double yy=sqrt(1-xx*xx);
                printf("%.6lf %.6lf\n",xx,yy);
            }
        }
    }
    return 0;
}