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

var t,i:byte;
    n:longint;
    s,s1:string;
    j,k,q,dd:byte;
    kt:boolean;

function divide(s:string;l,l1:byte):boolean;
var code:integer; du,x:longint; st:string;
    i,t:byte;
begin
     t:=l mod 4;
     if t=0 then t:=4;
     st:=copy(s,1,t);
     val(st,du,code);
     du:=du mod n;
     for i:=1 to (l-t) div 4 do
     begin
          st:=copy(s,t+i*4-3,4);
          val(st,x,code);
          du:=(du*10000+x) mod n;
     end;
     divide:=(du=0);
end;

begin
     readln(t);
     for i:=1 to t do
     begin
          readln(n);
          if n mod 5 = 0 then
          begin
               writeln(-1);
               continue;
          end;
          kt:=false;
          str(n,s1);
          dd:=length(s1);
          for j:=dd to 200 do
          begin
               for k:=0 to j do
               begin
                    s:='';
                    for q:=1 to k do s:=s+'8';
                    for q:=k+1 to j do s:=s+'6';
                    if divide(s,j,dd) then
                    begin
                         writeln(s);
                         kt:=true;
                         break;
                    end;
               end;
               if kt then break;
          end;
     end;
end.

Code mẫu của happyboy99x

#include<cstdio>

int pow10[201];

void testcase() {
    int x; scanf("%d", &x); x *= 9;
    pow10[0] = 1;
    for(int i = 1; i <= 200; ++i)
        pow10[i] = pow10[i-1] * 10 % x;
    for(int nd = 1; nd <= 200; ++nd)
        for(int n8 = 0; n8 <= nd; ++n8)
            if((8*pow10[nd] - 2*pow10[nd-n8] - 6) % x == 0) {
                for(int i = 0; i < n8; ++i) printf("8");
                for(int i = 0; i < nd - n8; ++i) printf("6");
                printf("\n");
                return;
            }
    printf("-1\n");
}

int main() {
    int tc; scanf("%d", &tc);
    while(tc--) testcase();
    return 0;
}

Code mẫu của ladpro98

program luckynum;
uses    math;
const   fi='';
        fo='';

type    bigNum = String;
var     xau:array[0..20400] of string;
        so6,so8:array[0..201] of string;
        inp,oup:text;
        t,n,i:longint;
        sochia:array[1..21] of longint;
function bigmod(a:bignum;b:longint):longint;
var     i,hold:longint;
begin
        hold:=0;
        for i:=1 to length(a) do
                hold:=(ord(a[i])-48+hold*10) mod b;
        exit(hold);
end;

procedure init;
var     s:string;
        i,j:longint;
begin
        s:='';
        so6[0]:='';
        for i:=1 to 200 do
        begin
                s:=s+'6';
                so6[i]:=s;
        end;

        s:='';
        so8[0]:='';
        for i:=1 to 200 do
        begin
                s:=s+'8';
                so8[i]:=s;
        end;
        t:=1;
        for i:=1 to 200 do
        begin
                for j:=0 to i do
                begin
                        xau[t]:=concat(so8[j],so6[i-j]);
                        inc(t);
                end;
        end;


end;

function process(x:longint):bignum;
var     i:longint;
begin
        if (x mod 5) = 0 then exit('-1');
        for i:=1 to t-1 do
        begin
                if (bigmod(xau[i],x) = 0) then exit(xau[i]);
        end;
        exit('-1');

end;


procedure input;
var     i:longint;
begin
        assign(inp,fi);
        reset(inp);
        assign(oup,fo);
        rewrite(oup);
        readln(inp,n);
        for i:=1 to n do
        begin
                readln(inp,sochia[i]);
        end;
        for i:=1 to n do
        begin

                writeln(oup,process(sochia[i]));

        end;
        close(inp);
        close(oup);
end;


begin
        init;
        input;
end.

Code mẫu của RR

{$R+,Q+}
uses math;
const
  FINP='';
  FOUT='';
  MAXN=201;
  lt10:array[0..3] of longint=(1,10,100,1000);
type
  big=array[0..51] of longint;
var
  n,test,t,best,s8,s6:longint;
  a:array[1..MAXN] of longint;
  b:big;
  f1,f2:text;
procedure openF; inline;
begin
  assign(f1,FINP); reset(f1);
  assign(f2,FOUT); rewrite(f2);
end;
procedure closeF; inline;
begin
  close(f1); close(f2);
end;
procedure ans; inline;
var
  i:longint;
begin
  if best=MAXN then
    begin
      writeln(f2,-1);
      exit;
    end;
  for i:=1 to s8 do write(f2,8);
  for i:=1 to s6 do write(f2,6);
  writeln(f2);
end;
procedure inp; inline;
begin
  readln(f1,n);
end;
procedure trans(l:longint; var b:big); inline;
var
  i,now:longint;
begin
  fillchar(b,sizeof(b),0);
  b[0]:=1; now:=0;
  for i:=l downto 1 do
    begin
      b[b[0]]:=b[b[0]]+a[i]*lt10[now];
      inc(now);
      if now=4 then
        begin
          now:=0;
          inc(b[0]);
        end;
    end;
  while (b[0]>0) and (b[b[0]]=0) do dec(b[0]);
end;
//Chia lay phan du
operator / (b:big; k:longint) c:longint;
var
  i,nho:longint;
begin
  nho:=0;
  for i:=b[0] downto 1 do
    nho:=(b[i]+nho*10000) mod k;
  c:=nho;
end;
procedure solve;
var
  i,j,ii:longint;
begin
  best:=MAXN;
  for i:=0 to MAXN-1 do
  for j:=0 to MAXN-1 do
    if (i>0) or (j>0) then
      begin
        if i+j>=best then continue;
        fillchar(a,sizeof(a),0);
        for ii:=1 to i do a[ii]:=8;
        for ii:=i+1 to i+j do a[ii]:=6;
        trans(i+j,b);
        if b/n=0 then
          begin
            best:=i+j;
            s8:=i; s6:=j;
          end;
      end;
end;
begin
  openF;
  read(f1,t);
  for test:=1 to t do
    begin
      inp;
      solve;
      ans;
    end;
  closeF;
end.

Code mẫu của hieult

#include <stdio.h>
main()
{
int n,a[20],t[20],n8[20],n6[20];
long b[20][201],c[20][201];
scanf("%d",&n);
for(int i=0;i<n;i++)
  scanf("%d",&a[i]);
for(int i=0;i<n;i++)
  {
  t[i]=0;
  b[i][0]=0;
  c[i][0]=0;
  b[i][1]=1;
  c[i][1]=1;
  for(int j=2;j<=200;j++)
    {
    b[i][j]=b[i][j-1]*10%a[i];
    c[i][j]=(c[i][j-1]*10+1)%a[i];
    }
  for(int j=1;j<=200;j++)
    {
    for(int k=0;k<=j;k++)
      if((8*b[i][j-k+1]*c[i][k]+6*c[i][j-k])%a[i]==0)
        {
        n8[i]=k;
        n6[i]=j-k;
        t[i]=1;
        break;
        }
    if(t[i]==1) break;
    }
  }
for(int i=0;i<n;i++)
  {
  if(t[i]==0)
    printf("-1\n");
  else
    {
    for(int j=0;j<n8[i];j++)
      printf("8");
    for(int j=0;j<n6[i];j++)
      printf("6");
    printf("\n");
    }
  }
}

Code mẫu của ll931110

{$MODE DELPHI}
Program LUCKYNUM;
Const
  input  = '';
  output = '';
  maxn = 10000;
  maxk = 200;
Var
  fi,fo: text;
  d6,d8: array[0..maxn] of integer;
  n,i,t: integer;

Procedure openfile;
Begin
  Assign(fi, input);
    Reset(fi);

  Assign(fo, output);
    Rewrite(fo);
End;

Procedure solve;
Var
  st,res,tmp,curr: integer;
  e6,e8,i,k: integer;
Begin
  res:= maxn;

  Fillchar(d8, sizeof(d8), 0);
  st:= 0;
  tmp:= st;

  Repeat
    Fillchar(d6, sizeof(d6), 0);
    d6[tmp]:= d8[tmp];
    If tmp = 0 then
      If (res > d8[tmp]) and (d8[tmp] <> 0) then
        Begin
          res:= d8[tmp];
          e6:= 0;
          e8:= d8[tmp];
        End;

    curr:= tmp;
    Repeat
      k:= (curr * 10 + 6) mod n;
      If d6[k] = 0 then d6[k]:= d6[curr] + 1 else break;
      curr:= k;
    Until false;

    If (d6[0] < res) and (d6[0] <> 0) then
      Begin
        res:= d6[0];
        e8:= d8[tmp];
        e6:= d6[0] - e8;
      End;

    tmp:= (st * 10 + 8) mod n;
    If d8[tmp] = 0 then d8[tmp]:= d8[st] + 1 else break;
    If d8[tmp] > res then break;
    st:= tmp;
  Until false;

  If res > maxk then writeln(fo, -1) else
    Begin
      For i:= 1 to e8 do write(fo, 8);
      For i:= 1 to e6 do write(fo, 6);
      Writeln(fo);
    End;
End;

Procedure closefile;
Begin
  Close(fo);
  Close(fi);
End;

Begin
  openfile;

  Readln(fi, t);
  For i:= 1 to t do
    Begin
      Readln(fi, n);
      solve;
    End;

  closefile;
End.

Code mẫu của skyvn97

#include<stdio.h>
#define MAX   222
int x;
int mod[MAX];
int m10[MAX];
int im,jm;
int t,c;
void process(void) {
    scanf("%d",&x);
    int i,j;
    mod[0]=0;
    m10[0]=1;
    for (i=1;i<=211;i=i+1) {
        mod[i]=(mod[i-1]*10+1)%x;
        m10[i]=(m10[i-1]*10)%x;
    }
    im=211;
    jm=211;
    for (i=0;i<=200;i=i+1)
        for (j=0;i+j<=200;j=j+1) {
            if (i+j==0) continue;
            if ((8*mod[i]*m10[j]+6*mod[j])%x!=0) continue;
            if (i+j>im+jm) continue;
            if (i+j==im+jm)
                if (i<im) {
                    im=i;
                    jm=j;
                    continue;
                }
            if (i+j<im+jm) {
                im=i; jm=j;
            }
        }
    if (im+jm>200) {
        printf("-1\n");
        return;
    }
    for (i=1;i<=im;i=i+1) printf("8");
    for (i=1;i<=jm;i=i+1) printf("6");
    printf("\n");
}
int main(void) {
    scanf("%d",&t);
    for (c=1;c<=t;c=c+1) process();
}