[Structure] FZU 2140 Forever 0.5, fzu2140

[Structure] FZU 2140 Forever 0.5, fzu2140

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Question: Enter N, which indicates there are N points,

Requirements:

1. The distance between any two points is less than or equal to 1.0.

2. The distance between each point and the origin is less than or equal to 1.0.

3. Distance between N points = 1.0

4. area formed by N points ≥ 0.5 ≤ 0.75

You can first take a vertex with the origin point and the other two points in the positive triangle on the unit circle.

Other points are on the arc BC.

Equals BC

And when N = 4, the fourth point must be in the BC midpoint.

#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;}