【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|1typedef 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;}

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