Lưu ý: Các code mẫu dưới đây chỉ mang tính tham khảo và có thể không AC được bài tập này

Code mẫu của flashmt

const fi='';
      fo='';
      maxn=10005;
      ep=0.000000001;
      pi2=6.283185307;
var n:longint;
    a:array[1..maxn] of real;
    re:real;

procedure rf;
var i:longint;
begin
     assign(input,fi); reset(input);
     readln(n);
     for i:=1 to n do read(a[i]);
     close(input);
end;

function calc(x:real):longint;
var i,z:longint; goc,cosin,tg,b,c,d:real;
begin
     goc:=0;
     for i:=1 to n do
     begin
          z:=i+1;
          if z=n+1 then z:=1;
          b:=x+a[i]; c:=x+a[z]; d:=a[i]+a[z];
          cosin:=(b*b+c*c-d*d)/(2*b*c);
          tg:=sqrt((1-sqr(cosin))/sqr(cosin));
          if cosin>=0 then goc:=goc+arctan(tg)
          else goc:=goc+pi2/2-arctan(tg);
     end;
     if abs(goc-pi2)<=ep then calc:=0
     else
         if goc>pi2 then calc:=1
         else calc:=-1;
end;

procedure pr;
var l,m,r:real; t:longint;
begin
     l:=0; r:=1000000000000000000000000;
     while r-l>=ep do
     begin
          m:=(l+r)/2;
          t:=calc(m);
          if t=0 then break;
          if t>0 then l:=m
          else r:=m;
     end;
     re:=m;
end;

procedure wf;
begin
     assign(output,fo); rewrite(output);
     writeln(re:0:3);
     close(output);
end;

begin
     rf;
     pr;
     wf;
end.

Code mẫu của happyboy99x

#include<cstdio>
#include<cmath>

const int N = 10000+5;
const double EPS = 1e-9;
double r[N];
int n;

inline double alpha(const double &a, const double &b, const double &c) {
    return acos((b*b+c*c-a*a)/(2*b*c));
}

int main() {
    scanf("%d",&n);for(int i=0;i<n;++i) scanf("%lf",r+i);
    double l = 0, h = 1e9;
    while(fabs(h-l) > EPS) {
        double r0 = (l+h)/2, angle = alpha(r[0]+r[n-1],r0+r[0],r0+r[n-1]);
        for(int i=1;i<n;++i) angle += alpha(r[i]+r[i-1],r0+r[i],r0+r[i-1]);
        if(angle < 2*M_PI) h = r0; else l = r0;
    }
    printf("%.3lf\n", (l+h)/2);
    return 0;
}

Code mẫu của ladpro98

program lem1;
uses    math;
const   fi='';
        maxn=10004;
        eps=1e-4;
        ok=360;
var     a:array[0..maxn] of extended;
        l,r,m,x,y,z,t,alpha:extended;
        n,i:longint;
        inp:text;

begin
        assign(inp,fi);reset(inp);
        readln(inp,n);
        for i:=1 to n do read(inp,a[i]);
        a[0]:=a[n];
        l:=0;r:=1e9;
        while (r-l>eps) do
        begin
                m:=(l+r)/2;
                alpha:=0;
                for i:=1 to n do
                begin
                        x:=m+a[i-1];
                        y:=m+a[i];
                        z:=a[i-1]+a[i];
                        t:=arccos((sqr(x)+sqr(y)-sqr(z))/(2*x*y));
                        alpha:=alpha+radtodeg(t);
                end;
                if alpha<ok then r:=m
                else    l:=m;
        end;
        write(m:0:3);
end.

Code mẫu của RR

//Written by Nguyen Thanh Trung
{$R+,Q+,N+}
{$Mode objFPC}
uses math;
const
  FINP          =       '';
  FOUT          =       '';
  MAXN          =       10111;
  eps           =       1e-12;
type
  real          =       extended;
var
  f1,f2         :       text;
  n             :       longint;
  r             :       array[1..MAXN] of real;

procedure openF;
    begin
      assign(f1,FINP); reset(f1);
      assign(f2,FOUT); rewrite(f2);
    end;
procedure closeF;
    begin
      close(f1);
      close(f2);
    end;
procedure inp;
    begin
      read(f1,n);
      for n:=1 to n do
        read(f1,r[n]);
    end;
procedure solve;
    var
      kq,left,right,mid,sum,a,b,c:real;
    begin
      r[n+1]:=r[1];
      left:=0; right:=0;
      for n:=1 to n do right+=r[n]*2; right/=pi*2;
      while left<right do
        begin
          mid:=(left+right)/2; sum:=0;
          for n:=1 to n do
            begin
              a:=r[n]+mid;
              b:=r[n+1]+mid;
              c:=r[n]+r[n+1];
              sum+=arccos((a*a+b*b-c*c)/2/a/b);
            end;
          if abs(sum-2*pi)<eps then begin writeln(f2,mid:0:3); exit; end;
          if sum<2*pi then
            begin
              kq:=mid;
              right:=mid-eps;
            end
          else left:=mid+eps;
        end;
      writeln(f2,kq:0:3);
    end;

begin
  openF;
  inp;
  solve;
  closeF;
end.

Code mẫu của hieult

#include <stdio.h>
//#include <conio.h>
#include <math.h>
#define pi 3.1415926535898
long long a[10003],n;
double Bcos(double t)
{
double T=0;
for(int i=1;i<=n;i++)
  T=T+acos(1-2*a[i]*a[i+1]/((a[i]+t)*(a[i+1]+t)));
return T;
}
main()
{
//freopen("LEM1.in","r",stdin);
scanf("%lld",&n);
for(int i=1;i<=n;i++)  
  scanf("%lld",&a[i]);
  a[n+1] = a[1];
double u = 0.01,v = 1000000000;
while(v-u>0.0000001)
{
     double r = (u+v)/2;
     if(Bcos(r)<2*pi)
        v=r;
     else u = r;
}
printf("%.3lf",u); 
//getch();
}

Code mẫu của ll931110

{$N+}
program LEM1;
uses math;
const
  input  = '';
  output = '';
  maxn = 10000 + 2;
  eps = 0.00000000001;
  eps2 = 0.0000000001;
  maxv = 1000000000;
var
  n: integer;
  a: array[1..maxn] of extended;
  res: extended;

procedure init;
var
  f: text;
  i: integer;
begin
  assign(f, input);
    reset(f);

  readln(f, n);
  for i := 1 to n do read(f, a[i]);
  a[n + 1] := a[1];

  close(f);
end;

function angle(x,y,z: extended): extended;
var
  t: double;
begin
  t := (y * y + z * z - x * x) / (2 * y * z);
  angle := arccos(t);
end;

procedure solve;
var
  inf,sup,med: extended;
  anl,tmp: extended;
  i: integer;
begin
  inf := 0;
  sup := 2 * maxv;
  tmp := 2 * pi;

  repeat
    med := (inf + sup) / 2;
    anl := 0;

    for i := 1 to n do
      anl := anl + angle(a[i] + a[i + 1],med + a[i],med + a[i + 1]);

    if abs(anl - tmp) < eps then
      begin
        res := med;
        exit;
      end;

    if anl < tmp then sup := med else inf := med;
  until sup - inf < eps2;

  res := (inf + sup) / 2;
end;

procedure printresult;
var
  f: text;
begin
  assign(f, output);
    rewrite(f);
    writeln(f, res:0:3);
  close(f);
end;

begin
  init;
  solve;
  printresult;
end.

Code mẫu của skyvn97

#include<stdio.h>
#include<math.h>
#define MAX   10101
#define INF   1e8
#define EPS   1e-5
#define PI   acos(-1.0)
double r[MAX];
int n,i;
double sa;
double t,b,m;
double angleA(double a,double b,double c) {
    return (acos((b*b+c*c-a*a)/(2*b*c)));
}
int main(void) {
    scanf("%d",&n);
    for (i=1;i<=n;i=i+1)
        scanf("%lf",&r[i-1]);
    b=0;t=INF;
    while (t-b>EPS) {
        m=(t+b)/2;
        sa=0;
        for (i=0;i<n;i=i+1)
            sa=sa+angleA(r[i%n]+r[(i+1)%n],m+r[i%n],m+r[(i+1)%n]);
        if (sa<2*PI) t=m; else b=m;     
    }       
    printf("%.3lf",b);
}

Code mẫu của khuc_tuan

uses math;
var
   n, i, kk, j,t : integer;
   r : array[1..10000] of double;
   xx,total, left, right, mid : double;
   ok : boolean;

function tinh(a,b,c : double) : double;
var
   xx : double;
begin
     xx :=  (b*b+c*c-a*a)/2/b/c;
     if(xx>=1-1e-8) then xx := 1;
     if(xx<=-1+1e-8) then xx := -1;
     tinh := arccos(xx);
end;

begin
      read(n);
      for i:=1 to n do read(r[i]);
      //for i:=1 to n do r[i] := 100000;
      left := 0;
      right := 1e9;
      for kk:=1 to 50 do begin
          mid := (left+right)/2;
          total := 0;
          ok := true;
          for i:=1 to n do begin
              t := i-1;
              if t=0 then t := n;
              j := i mod n + 1;

              xx :=  tinh( r[i]+r[j], r[i]+mid, r[j]+mid) ;
              //if( tinh(r[t]+r[i],r[t]+mid,r[i]+mid) + xx <= tinh( r[t]+r[j], r[t]+mid,r[j]+mid)-1e-8) then ok := false;
              total := total + xx;
          end;
          //if not ok then right := mid else
          if(total >= 2*3.1415926535) then left := mid
          else right := mid;
          {writeln( ok);
          writeln( total); }
      end;
      writeln( left:0:3);
end.