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 base=100000000;
      digit=8;
type bignum=array[0..875] of longint;
var num,re,temp:bignum;

procedure plus(var c:bignum;a,b:bignum);
var i,max,mem:longint; 
begin
     mem:=0;
     if a[0]>b[0] then max:=a[0] else max:=b[0];
     for i:=1 to max do
     begin
          c[i]:=a[i]+b[i]+mem;
          if c[i]<base then mem:=0
          else
          begin
               c[i]:=c[i]-base; mem:=1;
          end;
     end;
     if mem>0 then
     begin
          max:=max+1;
          c[max]:=mem;
     end;
     c[0]:=max;
end;

procedure plus1(var c:bignum;a,b,d:bignum);
var i,max,mem:longint; 
begin
     mem:=0;
     if a[0]>b[0] then max:=a[0] else max:=b[0];
     if max<d[0] then max:=d[0];
     for i:=1 to max do
     begin
          c[i]:=a[i]+b[i]+d[i]+mem;
          if c[i]<base then mem:=0
          else
          begin
               mem:=c[i] div base; c[i]:=c[i] mod base; 
          end;
     end;
     if mem>0 then
     begin
          max:=max+1;
          c[max]:=mem;
     end;
     c[0]:=max;
end;

procedure multi(var c:bignum;a:bignum;b:longint);
var i,max,mem:longint; 
begin
     mem:=0; max:=a[0];
     for i:=1 to max do
     begin
          c[i]:=a[i]*b+mem;
          if c[i]<base then mem:=0
          else
          begin
               mem:=c[i] div base; c[i]:=c[i] mod base; 
          end;
     end;
     if mem>0 then
     begin
          max:=max+1;
          c[max]:=mem;
     end;
     c[0]:=max;
end;

procedure mul2(var c:bignum;a:bignum;b:longint);
var i,max,mem:longint; 
begin
     mem:=0; max:=a[0];
     for i:=1 to max do
     begin
          c[i]:=a[i] shl b+mem;
          if c[i]<base then mem:=0
          else
          begin
               mem:=1; c[i]:=c[i]-base; 
          end;
     end;
     if mem>0 then
     begin
          max:=max+1;
          c[max]:=mem;
     end;
     c[0]:=max;
end;

procedure multi1(var c:bignum;a:bignum;b:longint;d:bignum);
var i,max,mem:longint; 
begin
     mem:=0; 
     if a[0]>=d[0] then max:=a[0] else max:=d[0];
     for i:=1 to max do
     begin
          c[i]:=a[i]*b+d[i]+mem;
          if c[i]<base then mem:=0
          else
          begin
               mem:=c[i] div base; c[i]:=c[i] mod base; 
          end;
     end;
     if mem>0 then
     begin
          max:=max+1;
          c[max]:=mem;
     end;
     c[0]:=max;
end;

procedure mul21(var c:bignum;a:bignum;b:longint;d:bignum);
var i,max,mem:longint; 
begin
     mem:=0; 
     if a[0]>=d[0] then max:=a[0] else max:=d[0];
     for i:=1 to max do
     begin
          c[i]:=a[i] shl b+d[i]+mem;
          if c[i]<base then mem:=0
          else
          begin
               mem:=c[i] div base; c[i]:=c[i] mod base; 
          end;
     end;
     if mem>0 then
     begin
          max:=max+1;
          c[max]:=mem;
     end;
     c[0]:=max;
end;

procedure star;
var i:longint;
begin
     multi1(re,re,5,num);
     multi(num,num,3);
end;

procedure rf;
var c:char;
begin
     num[0]:=1; re[0]:=1; re[1]:=1; num[1]:=1;
     while not eoln do
     begin
          read(c);
          if c='P' then continue;
          if c='*' then star
          else
          begin
               if c='L' then mul2(re,re,1) else mul21(re,re,1,num);
          end;
     end;
end;

procedure wf;
var i,j,t:longint; s:string;
begin
     for i:=re[0] downto 1 do
     begin
          if i<re[0] then
          begin
               str(re[i],s);
               t:=length(s);
               for j:=t+1 to digit do write(0);
          end;
          write(re[i]);
     end;
end;

begin
     rf;
     wf;
end.

Code mẫu của ladpro98

#include <bits/stdc++.h>
const int MOD = 100000000;
const int N = 10010;
using namespace std;
typedef vector<int> big;
char s[N];

big operator + (big a, big b) {
    big 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 % MOD);
        carry /= MOD;
    }
    if (carry) c.push_back(carry);
    return c;
}

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

void print(big a) {
    printf("%d", a[a.size() - 1]);
    for(int i = a.size() - 2; i >= 0; i--)
        printf("%08d", a[i]);
    printf("\n");
}

int main() {
    scanf("%s", &s);
    int n = strlen(s);
    big F; F.push_back(1);
    big aster; aster.push_back(1);
    for(int i = 0; i < n; i++) {
        if (s[i] == 'L') F = F + F; else
        if (s[i] == 'R') F = F + F + aster; else
        if (s[i] == '*') {
            F = (F * 5) + aster;
            aster = aster * 3;
        }
    }
    print(F);
    return 0;
}

Code mẫu của RR

//Written by RR
{$ifdef rr}
  {$r+,q+}
  {$mode objfpc}
  {$inline off}
{$else}
  {$r-,q-}
  {$mode objfpc}
  {$inline on}
{$endif}

uses math;
const
  FINP          =       '';
  FOUT          =       '';
  scs           =       400;
  base          =       1000000000000000000;
  lbase         =       18;
type
  big           =       array[0..scs] of qword;
var
  f1,f2         :       text;
  s             :       ansistring;
  n             :       longint;
  skn,val       :       big;

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
      readln(f1,s);
      n:=length(s);
    end;
procedure mul2(var a:big); inline;
    var
      i,nho:longint;
    begin
      nho:=0;
      for i:=1 to a[0] do
        begin
          a[i]:=a[i]<<1+nho;
          if a[i]<base then nho:=0
          else begin nho:=1; dec(a[i],base); end;
        end;
      if nho>0 then
        begin
          inc(a[0]);
          a[a[0]]:=nho;
        end;
    end;
procedure add(var a,b:big); inline;
    var
      i,nho:longint;
    begin
      nho:=0;
      if b[0]>a[0] then a[0]:=b[0];
      for i:=1 to a[0] do
        begin
          a[i]:=a[i]+b[i]+nho;
          if a[i]<base then nho:=0
          else begin nho:=1; dec(a[i],base); end;
        end;
      if nho>0 then
        begin
          inc(a[0]);
          a[a[0]]:=nho;
        end;
    end;
procedure mul3(var a:big); inline;
    var
      i,nho:longint;
    begin
      nho:=0;
      for i:=1 to a[0] do
        begin
          a[i]:=a[i]<<1+a[i]+nho;
          if a[i]<base then nho:=0
          else begin nho:=a[i] div base; a[i]:=a[i] mod base; end;
        end;
      if nho>0 then
        begin
          inc(a[0]);
          a[a[0]]:=nho;
        end;
    end;
procedure mul5(var a:big); inline;
    var
      i,nho:longint;
    begin
      nho:=0;
      for i:=1 to a[0] do
        begin
          a[i]:=a[i]<<2+a[i]+nho;
          if a[i]<base then nho:=0
          else begin nho:=a[i] div base; a[i]:=a[i] mod base; end;
        end;
      if nho>0 then
        begin
          inc(a[0]);
          a[a[0]]:=nho;
        end;
    end;
procedure solve;
    var
      i:longint;
    begin
      skn[0]:=1; skn[1]:=1;
      val[0]:=1; val[1]:=1;
      for i:=1 to n do
        case s[i] of
          'L': mul2(val);
          'R': begin mul2(val); add(val,skn); end;
          '*': begin mul5(val); add(val,skn); mul3(skn); end;
        end;
    end;
procedure print(var a:big);
    var
      s:string[20];
      i:longint;
    begin
      write(f2,a[a[0]]);
      for i:=a[0]-1 downto 1 do
        begin
          str(a[i],s);
          while length(s)<lbase do s:='0'+s;
          write(f2,s);
        end;
      writeln(f2);
    end;

begin
  openF;
  inp;
  solve;
  print(val);
  closeF;
end.

Code mẫu của hieult

//#include <conio.h>
#include <stdio.h>
#include <string.h>
#define du 100000000


struct so
{
    int scs,a[2000];
};

so gt,sl;
char s[10001];

void print(so A)
{
     int n = A.scs;
     printf("%d",A.a[n]);
     for(int i = n-1;i>=1;i--)
         printf("%08d",A.a[i]);
} 
         /* Tinh Tong */
void sum(so &tong, so b)
{
     so A = tong;

     if(A.scs > b.scs) for(int i = b.scs+1; i <= A.scs; i++) b.a[i] = 0;
     if(b.scs > A.scs) {for(int i = A.scs+1; i <= b.scs; i++) A.a[i] = 0; tong.scs = b.scs;} 
     int nho = 0;
     for(int i = 1; i <= tong.scs ; i++)
     {
             tong.a[i] = (A.a[i] + b.a[i] + nho)%du;
             nho = (A.a[i] + b.a[i] + nho)/du;
             //printf("%c ",a[i]);
     }    
     if(nho > 0){ tong.scs++ ; tong.a[tong.scs] = nho ;}
}

void tich(so &f,int k)
{
        int nho = 0,t;
        for(int j = 1;j<=f.scs;j++)
        {
            t = f.a[j];    
            f.a[j] = (t*k+nho)%du;
            nho = (t*k+nho)/du;
        }
        if(nho>0) f.a[++f.scs] = nho;
}
int main()
{  
    FILE* f;
    //f = fopen("SETNJA.in","wt");
   // for(int i = 1;i<=10000;i++)
       //fprintf(f,"*");
    //fclose(f);
    //freopen("SETNJA.in","r",stdin);
    scanf("%s",s);
    gt.scs = 1; gt.a[1] = 1;
    sl.scs = 1; sl.a[1] = 1;
    //printf("%d\n",strlen(s));
    for(int i = 0;i<strlen(s);i++)
    {
        if(s[i] == 'L')
            sum(gt,gt);
        else if(s[i]=='R')
        {
             sum(gt,gt);
             sum(gt,sl);
        }
        else if(s[i]=='*')
        {
             tich(gt,5);
             sum(gt,sl);
             tich(sl,3);
        }

    }
    print(gt);
        //getch();
}

Code mẫu của ll931110

{$inline on}
{$MODE DELPHI}
{$H+}
program SETNJA;
const
  input  = '';
  output = '';
  base = 100000000;
  maxd = 900;
type
  arr = array[1..maxd] of longint;
var
  m,res: arr;
  s: string;

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

  readln(f, s);

  close(f);
end;

procedure mul1(k: integer);inline;
var
  i: integer;
begin
  for i := 1 to maxd do res[i] := res[i] * k;
  for i := 1 to maxd - 1 do if res[i] > base then
    begin
      res[i + 1] := res[i + 1] + res[i] div base;
      res[i] := res[i] mod base;
    end;
end;

procedure mul2(k: integer);inline;
var
  i: integer;
begin
  for i := 1 to maxd do m[i] := m[i] * k;
  for i := 1 to maxd - 1 do if m[i] > base then
    begin
      m[i + 1] := m[i + 1] + m[i] div base;
      m[i] := m[i] mod base;
    end;
end;


procedure ad2;inline;
var
  i: integer;
begin
  for i := 1 to maxd do res[i] := res[i] + m[i];
  for i := 1 to maxd - 1 do if res[i] > base then
    begin
      inc(res[i + 1]);
      res[i] := res[i] - base;
    end;
end;

procedure solve;inline;
var
  i: integer;
begin
  fillchar(res, sizeof(res), 0);
  res[1] := 1;

  fillchar(m, sizeof(m), 0);
  m[1] := 1;

  for i := 1 to length(s) do
    case s[i] of
      'L': mul1(2);
      'R':
        begin
          mul1(2);
          ad2;
        end;
      '*':
        begin
          mul1(5);
          ad2;
          mul2(3);
        end;
    end;
end;

procedure printresult;inline;
var
  f: text;
  st: string;
  i,j,k: integer;
begin
  assign(f, output);
    rewrite(f);

  i := maxd;
  while res[i] = 0 do dec(i);

  write(f, res[i]);
  for j := i - 1 downto 1 do
    begin
      str(res[j],st);
      for k := 1 to 8 - length(st) do write(f, 0);
      write(f, res[j]);
    end;

  close(f);
end;

begin
  init;
  solve;
  printresult;
end.

Code mẫu của khuc_tuan

s = raw_input()
t = 1
k = 1
for c in s:
    if c=='L':
        t *= 2
    if c=='R':
        t = 2 * t + k
    if c=='*':
        t = 5 * t + k
        k *= 3
print t