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>
const int BASE = 10000;
const int N = 1010;
using namespace std;

typedef vector<int> BigInt;
BigInt F[N];

BigInt operator + (BigInt a, BigInt b) {
    BigInt c; int carry = 0;
    for(int i = 0; i < a.size() || i < b.size(); i++) {
        if (i < a.size()) carry += a[i];
        if (i < b.size()) carry += b[i];
        c.push_back(carry % BASE);
        carry /= BASE;
    }
    if (carry) c.push_back(carry);
    return c;
}

BigInt operator * (BigInt a, BigInt b) {
    BigInt c(a.size() + b.size() + 2, 0);
    for(int i = 0; i < a.size(); i++) {
        int carry = 0;
        for(int j = 0; j < b.size(); j++) {
            int k = i + j;
            c[k] += a[i] * b[j] + carry;
            carry = c[k] / BASE;
            c[k] %= BASE;
        }
        if (carry) c[i + b.size()] += carry;
    }
    while (*c.rbegin() == 0 && c.size() > 0)
        c.erase(c.begin() + c.size() - 1);
    return c;
}

void Print(BigInt a) {
    int L = a.size();
    printf("%d", a[L - 1]);
    for(int i = L - 2; i >= 0; i--)
        printf("%04d", a[i]);
    printf("\n");
}

int main()
{
    int i, j, k, l, U, D, L, R, n;
    F[1].push_back(1); F[2].push_back(2);
    for(i = 3; i < N; i++) F[i] = F[i - 1] + F[i - 2];
    BigInt res;
    #define DOWN 0
    while (scanf("%d", &n) == 1) {
        res = F[2];
        for(i = 0; i < 2; i++) for(j = 0; j < 2; j++)
        for(k = 0; k < 2; k++) for(l = 0; l < 2; l++) {
            U = D = L = R = n;
            if (i == DOWN) L--; else U--;
            if (j == DOWN) R--; else U--;
            if (k == DOWN) R--; else D--;
            if (l == DOWN) L--; else D--;
            res = res + (F[U] * F[D] * F[L] * F[R]);
        }
        Print(res);
    }
    return 0;
}

Code mẫu của RR

//Wishing myself a happy lunar new year with a lot of accept solutions
//Written by Nguyen Thanh Trung
{$R-,Q-}
uses math;
const
  FINP='';
  FOUT='';
  MAXN=100;
  MAXM=1000;
  oo=1000000000;
type
  bigNum=array[0..MAXN] of longint;
var
  f1,f2:text;
  fib,fib2,fib3:array[-1..MAXM] of bigNum;
  a,tg,kq:bigNum;
  n:longint;
procedure openF;
begin
  assign(f1,FINP); reset(f1);
  assign(f2,FOUT); rewrite(f2);
end;
procedure closeF;
begin
  close(f1); close(f2);
end;
operator +(a,b:bigNum) c:bigNum;
var
  nho,i:longint;
begin
  nho:=0;
  fillchar(c,sizeof(c),0); c[0]:=max(a[0],b[0]);
  for i:=MAXN downto MAXN-c[0]+1 do
    begin
      c[i]:=a[i]+b[i]+nho;
      nho:=c[i] div oo;
      c[i]:=c[i] mod oo;
    end;
  if nho>0 then
    begin
      inc(c[0]);
      c[MAXN-c[0]+1]:=nho;
    end;
end;
operator *(a,b:bigNum) c:bigNum;
var
  nho,i,j,k:longint;
  temp:int64;
begin
  fillchar(c,sizeof(c),0);
  if (a[0]=0) or (b[0]=0) then exit;
  if (a[0]=1) and (a[MAXN]=0) then exit;
  if (b[0]=1) and (b[MAXN]=0) then exit;
  k:=0;
  for i:=MAXN downto MAXN-a[0]+1 do
    begin
      fillchar(tg,sizeof(tg),0); nho:=0;
      tg[0]:=b[0]+k;
      for j:=MAXN downto MAXN-b[0]+1 do
        begin
          temp:=int64(b[j])*a[i]+nho;
          nho:=temp div oo;
          tg[j-k]:=temp mod oo;
        end;
      if nho>0 then
        begin
          inc(tg[0]);
          tg[MAXN-tg[0]+1]:=nho;
        end;
      while (tg[0]>0) and (tg[MAXN-tg[0]+1]=0) do dec(tg[0]);
      c:=c+tg;
      inc(k);
    end;
end;
procedure init;
var
  i:longint;
begin
  fib[0,0]:=1; fib[0,MAXN]:=1;
  fib[1,0]:=1; fib[1,MAXN]:=1;
  for i:=2 to MAXM do
    fib[i]:=fib[i-1]+fib[i-2];
  for i:=0 to MAXM do
    fib2[i]:=fib[i]*fib[i];
  for i:=0 to MAXM do
    fib3[i]:=fib[i]*fib2[i];
end;
procedure print(a:bigNum);
var
  i:longint;
  s:ansistring;
begin
  if a[0]=0 then begin writeln(f2,0); exit; end;
  write(f2,a[MAXN-a[0]+1]);
  for i:=MAXN-a[0]+2 to MAXN do
    begin
      str(a[i],s);
      while length(s)<9 do s:='0'+s;
      write(f2,s);
    end;
  writeln(f2);
end;
procedure solve;
begin
  fillchar(kq,sizeof(kq),0);
  kq[0]:=1; kq[MAXN]:=2;
  a[0]:=1; a[MAXN]:=16;
  kq:=kq+a*fib[n-2]*fib3[n-2];
  a[MAXN]:=64;
  kq:=kq+a*fib[n-3]*fib3[n-2];
  a[MAXN]:=16;
  kq:=kq+a*fib[n-4]*fib3[n-2];
  a[MAXN]:=80;
  kq:=kq+a*fib2[n-3]*fib2[n-2];
  a[MAXN]:=32;
  kq:=kq+a*fib[n-4]*fib[n-3]*fib2[n-2];
  a[MAXN]:=32;
  kq:=kq+a*fib3[n-3]*fib[n-2];
  a[MAXN]:=2;
  kq:=kq+a*fib[n-3]*fib3[n-3];
  a[MAXN]:=2;
  kq:=kq+a*fib2[n-4]*fib2[n-2];
  a[MAXN]:=12;
  kq:=kq+a*fib[n-4]*fib2[n-3]*fib[n-2];
end;
begin
  openF;
  init;
  while not eof(f1) do
    begin
      readln(f1,n);
      solve;
      print(kq);
    end;
  closeF;
end.

Code mẫu của hieult

#include <cstdio>
#include <iostream>
//#include <conio.h>
#define base 100000000

using namespace std;

struct solon
{
     int so;
     int a[200];
};

solon f[1010];
int n,r[10],mu2[10],rong;

solon tong (solon A,solon B)
{
      int du = 0;
      solon C;
      if(A.so<B.so)
      {
           C = A;
           A = B;
           B = C;
      }
      for(int i = B.so+1;i<=A.so;i++) B.a[i] = 0;
      C.so = A.so;
      for(int i = 1;i<=A.so;i++)
      {
          C.a[i] = (A.a[i]+B.a[i]+du)%base;
          du = (A.a[i]+B.a[i]+du)/base;
      }
      if(du>0) {C.so++; C.a[C.so] = du;}
      return C;
}

solon tichnho(solon A,long long k,int chuso)
{
      solon C;
      long long du = 0;
      C.so = A.so+chuso;
      for(int i = 1;i<=chuso;i++)
           C.a[i] = 0;
      for(int i = chuso+1;i<=chuso+A.so;i++)
      {
           C.a[i] = (A.a[i-chuso]*k+du)%base;
           du = (A.a[i-chuso]*k+du)/base;
      }
      if(du>0) {C.so++; C.a[C.so] = du;}
      return C;
}

solon tich(solon A,solon B)
{
      solon C;
      C.so = 1; C.a[1] = 0;
      for(int i = 1;i<=B.so;i++)
      {
           C = tong(C,tichnho(A,B.a[i],i-1));
      }
      return C;
}

void print(solon A)
{
      printf("%d",A.a[A.so]);
      for(int i = A.so-1;i>=1;i--)
          printf("%08d",A.a[i]);
}

int main()
{
      mu2[0] = 1; for(int i = 1;i<=10;i++) mu2[i]=mu2[i-1]*2;
      f[0].so = 1; f[0].a[1] = 1;
      f[1].so = 1; f[1].a[1] = 1;
      for(int i = 2;i<=1000;i++) f[i] = tong(f[i-1],f[i-2]);
      while(scanf("%d",&n)>0)
      {
           if(n==3) printf("196\n");
           else{
               solon KQ,nhan;int lan;
               KQ.so = 1; KQ.a[1] = 2;
               for(int i = 0;i<81;i++)
               {
                     int the = i;
                     for(int j = 1;j<=4;j++){
                          r[j] = the%3;
                          the/=3;
                     }
                     r[5] = r[1];
                     nhan.so = 1; nhan.a[1] = 1;
                     lan = 0;
                     for(int j = 1;j<=4;j++){
                          rong = n-2;
                          if(r[j]==2) rong--;
                          if(r[j+1]==1) rong--;
                          if(r[j]==0) lan++;
                          nhan = tich(nhan,f[rong]);
                     }
                     nhan = tichnho(nhan,mu2[lan],0);
                     KQ = tong(KQ,nhan);
               }
               print(KQ);
               printf("\n");
           }     
      }
}

Code mẫu của khuc_tuan

import java.io.*;
import java.util.*;
import java.math.*;
public class Main {
static int[][] luu = {{-2,-2,-2,-2,16},{-3,-2,-2,-2,64},{-3,-3,-2,-2,80},{-4,-2,-2,-2,16},{-3,-3,-3,-2,32},{-4,-3,-2,-2,32},{-3,-3,-3,-3,2},{-4,-3,-3,-2,12},{-4,-4,-2,-2,2},{1,1,1,1,2}};
public static void main(String[] args) {
BigInteger[] F = new BigInteger[10000];
F[0] = BigInteger.ONE;
F[1] = BigInteger.ONE;
for(int i=2;i<F.length;++i) {
F[i] = BigInteger.ZERO;
F[i] = F[i].add(F[i-1]);
F[i] = F[i].add(F[i-2]);
}
Scanner sc = new Scanner(System.in);
while(sc.hasNext()) {
int n = sc.nextInt();
if(n==3) System.out.println(196);
else {
BigInteger res = BigInteger.ZERO;
for(int i=0;i<luu.length;++i) {
BigInteger tmp = BigInteger.ONE;
for(int j=0;j<luu[i].length-1;++j) if(luu[i][j]<0) tmp = tmp.multiply(F[n+luu[i][j]]);
tmp = tmp.multiply(BigInteger.valueOf(luu[i][luu[i].length-1]));
res = res.add(tmp);
}
System.out.println(res);
}
}
}
}