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

Code mẫu của ladpro98

#include <bits/stdc++.h>
#define ii pair<int, int>
#define X first
#define Y second
const int N = 1505;
using namespace std;
int n;
int F[N][N], G[N][N];
ii a[N];

int area(int i, int j, int k)
    {return abs(a[i].X * (a[j].Y - a[k].Y) + a[j].X * (a[k].Y - a[i].Y) + a[k].X * (a[i].Y - a[j].Y));}

int main() {
    scanf("%d", &n);
    for(int i = 0; i < n; i++) scanf("%d %d", &a[i].X, &a[i].Y);
    for(int i = 0; i < n; i++) {
        int pos = i;
        for(int j = i + 1; j < n; j++) {
            while (pos < j && area(i, pos, j) <= area(i, pos + 1, j)) pos++;
            F[i][j] = area(i, pos, j);
        }
    }
    #define next(i) ((i) - 1 + n) % n
    for(int i = 0; i < n; i++) {
        int pos = i;
        for(int j = next(i); j != i; j = next(j)) {
            while (pos != j && area(i, pos, j) <= area(i, next(pos), j)) pos = next(pos);
            G[i][j] = area(i, pos, j);
        }
    }
    int res = 0;
    for(int i = 0; i < n; i++) for(int j = i + 1; j < n; j++)
        res = max(res, F[i][j] + G[i][j]);
    printf("%.1f", res / 2.0);
    return 0;
}

Code mẫu của RR

{$R+,Q+}
{$Mode objFPC}
uses math;
const
  FINP='';
  FOUT='';
var
  d,c:array[1..1501,1..1501] of longint;
  n:longint;
  x,y:array[1..3001] of longint;
  f1,f2:text;
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,x[n],y[n]);
end;
function dt(i,j,k:longint):longint;
begin
  exit(abs((x[j]-x[i])*(y[j]+y[i])
          +(x[k]-x[j])*(y[k]+y[j])
          +(x[i]-x[k])*(y[i]+y[k])));
end;
procedure solve;
var
  best,i,j,ii,temp:longint;
begin
  best:=0;
  for i:=1 to n-2 do
    begin
      ii:=i+1;
      for j:=i+2 to n-1 do
        begin
          c[i,j]:=0;
          repeat
             temp:=dt(ii,i,j);
             if (temp>=c[i,j]) then c[i,j]:=temp
             else break;
             inc(ii);
             if ii>j then break;
          until false;
          dec(ii);
        end;
    end;
  for i:=1 to n-2 do
    begin
      ii:=i+3;
      for j:=i+2 to n-1 do
        begin
          d[i,j]:=0;
          repeat
            temp:=dt(ii,i,j);
            if (temp>=d[i,j]) then d[i,j]:=temp
            else break;
            inc(ii);
            if ii>n then break;
          until false;
          dec(ii);
        end;
    end;
  for i:=1 to n-2 do
  for j:=i+2 to n-1 do
    best:=max(best,c[i,j]+d[i,j]);
  if best mod 2=0 then writeln(f2,best div 2,'.0')
  else writeln(f2,best div 2,'.5');
end;
begin
  openF;
  inp;
  solve;
  closeF;
end.

Code mẫu của hieult

#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<cassert>
#include<ctime>
#include<algorithm>
#include<iterator>
#include<iostream>
#include<cctype>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<list>
//#include<conio.h>
#define ep 0.000000001
#define maxn 10011
#define oo 1111111
#define mod 1000000000
#define TR(c, it) for(typeof((c).begin()) it=(c).begin(); it!=(c).end(); it++)
#define fi first 
#define se second
double const PI=4*atan(1);

using namespace std;

typedef pair<int, int> II;
typedef vector<int> VI;
typedef vector<II> VII;
typedef vector<VI> VVI;
typedef vector<VII> VVII;

struct Point{
    int x, y;
    Point(){};
    Point(int _x, int _y){
        x = _x;
        y = _y;
    }
};

Point a[maxn * 2];
int n;

int dientich(int i, int j, int k){
    int T = (a[i].x * a[j].y - a[j].x * a[i].y) +
            (a[j].x * a[k].y - a[k].x * a[j].y) +
            (a[k].x * a[i].y - a[i].x * a[k].y) ; 
    return abs(T);
}

int main(){
   // freopen("input.in","r",stdin); 
   // freopen("output.out","w",stdout);

    scanf("%d",&n);
    for(int i = 1; i <= n; i++){
        scanf("%d %d", &a[i].x, &a[i].y);
        a[n+i] = a[i];
    }

    int kq = -oo;

    for(int i = 1; i <= n; i++){
        int u = i + 1, v = i + 3;
        for(int j = i + 2; j <= n; j ++){
            while(dientich(i, j, u + 1) >= dientich(i, j, u) && u < j) u ++;
            while(dientich(i, v + 1, j) >= dientich(i, v, j) && v < n) v ++;
            kq = max(kq, dientich(i, j, u) + dientich(i, v, j));
        }
    }

    printf("%.1lf",kq * 1.0 / 2);
     // getch();
}

Code mẫu của khuc_tuan

#include <iostream>
#include <cmath>
using namespace std;

struct Point {
    int x, y;
};

int n;
Point p[1550];

inline int getS(Point a, Point b, Point c) {
    return abs((a.x-b.x) * (a.y+b.y) + (b.x-c.x) * (b.y+c.y) + (c.x-a.x) * (c.y+a.y));
}

int main() {
    scanf("%d", &n);
    for(int i=0;i<n;++i)
        scanf("%d%d", &p[i].x, &p[i].y);
    int a, b, i, j;
    int result = 0;
    for(a=0;a<n;++a) {
        i = (a+1) % n;
        j = (a+3) % n;
        for(b=a+2;b<n;++b) {
            int cur = getS(p[a],p[b],p[i]), tmp = getS(p[a],p[b],p[(i+1)%n]);
            while(tmp >= cur && (i+1)%n!=b) {
                i = (i+1) % n;
                cur = tmp;
                tmp = getS(p[a],p[b],p[(i+1)%n]);
            }
            cur = getS(p[a],p[b],p[j]);
            tmp = getS(p[a],p[b],p[(j+1)%n]);
            while(tmp >= cur && (j+1)%n!=a) {
                j = (j+1) % n;
                cur = tmp;
                tmp = getS(p[a],p[b],p[(j+1)%n]);
            }
            result >?= getS(p[a],p[b],p[i]) + getS(p[a],p[b],p[j]);
        }
    }
    if(result%2==0) cout << result / 2 << ".0" << endl;
    else cout << result / 2 << ".5" << endl;
    return 0;
}