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

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<utility>
#define fr(a,b,c) for (a=b;a<=c;a++)
#define frr(a,b,c) for (a=b;a>=c;a--)
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
#define F first
#define S second
#define oo 1000111222
using namespace std;

int n,m,p;
long long f[6][1000100],g[1000100],re;

int main()
{
    int i,j,s,t;
    long long u;
    cin >> n >> m;
    n>>=1;
    p=(n+1)/2;
    fr(j,0,9) f[1][j*(n%2?1:11)%m]++;
    if (p==1) g[0]=1;
    u=(n%2?1:10);
    fr(i,1,p-1)
    {
        u*=100;
       fr(s,0,m-1)
            if (f[i][s])
                fr(j,0,9)
                {
                    t=(u*j+j+s*10)%m;
                    f[i+1][t]+=f[i][s];
                    if (i==p-1 && !j) g[t]+=f[i][s];
                }
    }
    u*=10; u%=m;
    fr(i,0,m-1)
    {
        s=1LL*m-(u*i)%m;
        s%=m;
        re+=(f[p][i]-g[i])*f[p][s];
    }
    cout  << re << endl;
   return 0;
}

Code mẫu của ladpro98

program MBIPALIN;
uses    math;
const   fi='';
        maxB=trunc(1e6);
var     m,n,lim,i:longint;
        pow:array[0..18] of int64;
        f:array[0..maxB] of longint;
        res,B:int64;
        inp:text;

procedure back(i,forw,rev,t:longint);
var     v:int64;
        j:longint;
begin
        if i>lim then begin
                if odd(m) then
                v:=(forw div 10)*pow[m-lim+1] mod B
                else v:=forw*pow[m-lim] mod B;
                v:=(v+rev) mod B;
                if t=1 then inc(f[v])
                else    inc(res,f[(B*B-v*pow[m]) mod B]);
                exit;
        end;
        for j:=0 to 9 do
        back(i+1,forw*10+j,j*pow[i-1]+rev,t);
end;

begin
        assign(inp,fi);reset(inp);
        readln(inp,n,B);close(inp);
        m:=n div 2;lim:=(m+1) div 2;pow[0]:=1;
        for i:=1 to 18 do pow[i]:=(10*pow[i-1]) mod B;
        back(1,0,0,1);
        for i:=1 to 9 do
        back(2,i,i,2);
        write(res);
end.

Code mẫu của RR

//Written by RR
{$R-,Q-}
{$Mode objfpc}
{$inline on}

uses math;
const
  FINP          =       '';
  FOUT          =       '';
  MAXN          =       22;

var
  f1,f2         :       text;
  n,m,mid,now   :       longint;
  lt10          :       longint;
  res           :       int64;
  count         :       array[0..1000111] of longint;
  kq            :       array[1..MAXN] of longint;

procedure openF;
    begin
      assign(f1,FINP); reset(f1);
      assign(f2,FOUT); rewrite(f2);
    end;

procedure closeF;
    begin
      close(f1);
      close(f2);
    end;

procedure duyet(i:longint;update:boolean); inline;
    var
      j,save:longint;
    begin
      if i<=mid then
        begin
          for j:=0 to 9 do
            begin
              kq[i]:=j;
              save:=now;
              now:=(now*10+j) mod m;
              if i<n then duyet(i+1,update)
              else
                begin
                  if not update then inc(count[now])
                  else if kq[1]>0 then
                    inc(res,count[(m-(lt10*now) mod m) mod m]);
                end;
              now:=save;
            end;
        end
      else
        begin
          j:=kq[n-i+1];
          save:=now;
          now:=(now*10+j) mod m;
          if i<n then duyet(i+1,update)
          else
            begin
              if not update then inc(count[now])
              else if kq[1]>0 then
                  inc(res,count[(m-(int64(lt10)*now) mod m) mod m]);
            end;
          now:=save;
        end;
    end;

procedure solve;
    var
      i:longint;
    begin
      now:=0;
      duyet(1,false);
      lt10:=1;
      for i:=1 to n do
        lt10:=(lt10*10) mod m;

      now:=0;
      duyet(1,true);

      writeln(f2,res);
    end;

begin
  openF;
  read(f1,n,m);
  n:=n div 2; mid:=n div 2+n mod 2;
  solve;
  closeF;
end.

Code mẫu của hieult

#include <stdio.h>
//#include <conio.h>

long long du1[1000001],du2[1000001];

int main()
{
    int n,m;
    long long mu10[11];
    mu10[0]=1;
    for(int i = 1;i<=10;i++)
        mu10[i]=mu10[i-1]*10;
    scanf("%d %d",&n,&m);
    n = n/2;
    for(int i = 0;i<m;i++)
    {
        du1[i]=0;
        du2[i]= 0;
    }
    int k = (n+1)/2;
    for(int i  = 0;i<mu10[k];i++)
    {
        long long t = 0,h=i;
        if(n%2==0)
        {
             t = h*mu10[k];
             for(int j = 1;j<=k;j++)
             {
                 int r = h%10;
                 t = t+r*mu10[k-j];
                 h = h/10;
             }
        }
        else
        {
              t = h*mu10[k-1];
              h = h/10;
              for(int j = 2;j<=k;j++)
              {
                  int r = h%10;
                  t  = t+r*mu10[k-j];
                  h = h/10;
              }
        }
        du2[t%m]++;
        if(i>=mu10[k-1])
            du1[(t*(mu10[n]%m))%m]++;
    }
    long long KQ = 0;
    KQ = du1[0]*du2[0];
    for(int i = 1;i<m;i++)
        KQ = KQ+du1[i]*du2[m-i];
    printf("%lld",KQ);
   // getch();
}

Code mẫu của ll931110

{$MODE DELPHI}
Program MBIPALIN;
Const
  input  = '';
  output = '';
  maxm = 1000000;
Var
  pre,sup: array[0..maxm - 1] of int64;
  d: array[0..20] of integer;
  n,m,k: integer;

Procedure init;
Var
  f: text;
Begin
  Assign(f, input);
    Reset(f);
    Readln(f, n, m);
  Close(f);
End;

Function calc: int64;
Var
  t: int64;
  i: integer;
Begin
  t:= d[1];
  For i:= 2 to k do t:= t * 10 + d[i];
  calc:= t mod m;
End;

Procedure solve;
Var
  tmp,s: integer;
  t: int64;
Begin
  k:= n div 2;
  s:= k div 2;
  If odd(k) then inc(s);

  Fillchar(pre, sizeof(pre), 0);
  fillchar(sup, sizeof(sup), 0);
  Fillchar(d, sizeof(d), 0);

  While d[0] = 0 do
    Begin
      t:= calc;
      inc(sup[t]);
      If d[1] <> 0 then inc(pre[t]);

      tmp:= s;
      inc(d[tmp]);
      d[k + 1 - tmp]:= d[tmp];

      While d[tmp] > 9 do
        Begin
          d[tmp]:= d[tmp] - 10;
          d[k + 1 - tmp]:= d[tmp];

          dec(tmp);
          inc(d[tmp]);
          d[k + 1 - tmp]:= d[tmp];
        End;
    End;
End;

Procedure printresult;
Var
  res,supp,tmp: int64;
  i: integer;
  f: text;
Begin
  res:= 0;

  tmp:= 1;
  For i:= 1 to k do tmp:= tmp * 10;

  For i:= 0 to m - 1 do
    Begin
      supp:= i;
      supp:= (supp * tmp) mod m;

      supp:= m - supp;
      If supp = m then supp:= 0;

      res:= res + pre[i] * sup[supp];
    End;

  Assign(f, output);
    Rewrite(f);
    Writeln(f, res);
  Close(f);
End;

Begin
  init;
  solve;
  printresult;
End.

Code mẫu của khuc_tuan

import java.util.Scanner;

public class Main {
    static int n, m, dd;
    static int[] a;
    static int[] c0, c1;

    static void generate(int pos) {
        if (pos == (n + 1) / 2) {
            long r = 0;
            for (int i = 0; i < n; ++i)
                r = r * 10 + a[i];
            if (a[0] != 0)
                c0[(int) (dd * r % m)]++;
            c1[(int) (r % m)]++;
            return;
        }
        for (int cs = 0; cs < 10; ++cs) {
            a[pos] = a[n - 1 - pos] = cs;
            generate(pos + 1);
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        m = sc.nextInt();
        n /= 2;
        c0 = new int[m];
        c1 = new int[m];
        dd = 1;
        for (int i = 0; i < n; ++i)
            dd = dd * 10 % m;
        a = new int[n];
        generate(0);
        long r = 0;
        for (int i = 0; i < m; ++i)
            r = (r + c0[i] * (long) c1[i == 0 ? 0 : (m - i)]);
        System.out.println(r);
    }
}