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 p:array[1..7] of longint=(1,10,100,1000,10000,100000,1000000);
var n,i,j,l,m,x:longint;
    a,s,u:string;
    code:integer;
    b:array[1..200000] of longint;

procedure gen(z,i:longint);
var j:longint;
begin
     if i>z then
     begin
          m:=m+1; b[m]:=x;
          if z=7 then
          begin
               m:=m+1; b[m]:=x+10000000;
          end;
          exit;
     end;
     for j:=1 to 5 do
     begin
          x:=x+p[i]*(j*2-1);
          gen(z,i+1);
          x:=x-p[i]*(j*2-1);
     end;
end;

function chk(a:string):boolean;
var i:longint;
begin
     for i:=1 to length(a) do
         if a[i]<'2' then exit(false);
     exit(true);
end;

function kt(x:longint):boolean;
begin
     while x>0 do
     begin
          if not odd(x mod 10) then exit(false);
          x:=x div 10;
     end;
     exit(true);
end;

function chk2(x:longint):boolean;
var i:longint;
begin
     for i:=1 to m do
         if (b[i]<x) and kt(x-b[i]) then exit(true);
     exit(false);
end;

begin
     for i:=1 to 7 do
     begin
          x:=0; gen(i,1);
     end;
     while true do
     begin
          readln(s);
          if s='[END]' then break;
          if s='[CASE]' then
          begin
               readln(n);
               if odd(n) then inc(n);
               str(n,a);
               l:=length(a);
               if chk(a) or chk2(n) then
               begin
                    writeln(n); continue;
               end;
               for i:=l downto 2 do
                   if a[i]<'2' then
                   begin
                        u:=copy(a,1,i-1);
                        val(u,x,code);
                        if chk(u) or chk2(x) then
                        begin
                             write(u);
                             for j:=i to l do write(2);
                             writeln;
                             break;
                        end;
                   end;
          end;
     end;
end.

Code mẫu của ladpro98

#include <bits/stdc++.h>
const int M = 1000006;
using namespace std;
int a[M];
int m;

int main() {
    ios :: sync_with_stdio(0); 
    for(int i = 1; i <= 9; i += 2) a[++m] = i;
    int cut = 1;
    for(int i = 2; i <= 8; i++) {
        int tmp = m;
        for(int j = cut; j <= tmp; j++)
            for(int k = 1; k <= 9; k += 2) 
                a[++m] = 10 * a[j] + k;
        cut = tmp;
    }
    string ins; int x;
    while (1) {
        cin >> ins;
        if (ins[1] == 'E') break;
        cin >> x;
        int j = lower_bound(a + 1, a + 1 + m, x) - a;
        int res = a[m] + a[m];
        for(int i = 1; i <= j; i++) {
            while (j > i && a[i] + a[j - 1] >= x) j--;
            if (a[i] + a[j] >= x) res = min(res, a[i] + a[j]);
        }
        cout << res << endl;
    }
    return 0;
}

Code mẫu của RR

#include <iostream>
#include <algorithm>

#define FOR(i,a,b)  for(int i=a; i<=b; i++)
#define FORD(i,a,b) for(int i=a; i>=b; i--)
using namespace std;

int n,a[500111],sl;
char s[20];

void dequy(int i,int gh,int now) {
    for(int u=1; u<=9; u+=2) {
        int x=now*10+u;
        if (i<gh) dequy(i+1,gh,x);
        else a[++sl]=x;
    }
}

void init() {
    FOR(l,1,8) dequy(1,l,0);
}

void solve() {
    int u=lower_bound(a+1,a+sl+1,n)-a-1; if (u<sl) u++;

    int res=a[u]+a[u]+1;
    a[0]=-a[sl]-1;
    int j=u;
    FOR(i,1,u) {
        if (i>j) break;
        if (a[i]>res) break;
        while (a[i]+a[j-1]>=n) j--;
        while (a[i]+a[j]>res) j--;
        if (a[i]+a[j]>=n) res=min(res,a[i]+a[j]);
    }
    printf("%d\n",res);
}

int main() {
    init();
    scanf("%s\n",&s);
    while (s[1]=='C') {
        scanf("%d",&n);
        solve();
        scanf("%s\n",&s);
    }
    return 0;
}

Code mẫu của hieult

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

    int k[] = {0,1,3,5,7,9};
    int mu5[10],a[500000],mu10[10],b[100000];

int main()
{
    //freopen("TOTALODD.in","r",stdin);

        mu5[0] = 1; mu10[0] = 1;
    for(int i = 1;i<=9;i++)
    {
        mu5[i] = mu5[i-1]*5;
        mu10[i] = mu10[i-1]*10;
    }
    for(int i = 1;i<=5;i++)
    {
        a[i] = k[i];
        b[i] = k[i];
    }
    int chay = 5;
    for(int i = 2;i<=8;i++)
    {
         int t,K;
         for(int j = 1;j<=mu5[i];j++)
         {
              t = a[(j-1)/mu5[i-1]+1];
              a[chay+j] = t*mu10[i-1]+b[(j-1)%mu5[i-1]+1];
         }
         if(i!=8)
              for(int j = 1;j<=mu5[i];j++)
                   b[j] = a[chay+j];
         chay = chay+mu5[i];
    }
    char s[20];
    int n;
    //printf("%d",chay);
    while(scanf("%s",s)>0 && s[1] != 'E')
    {
         scanf("%d",&n);
         int u,v,r,min = 100000001;
         if(n>=a[chay])
             u = chay;
         else
         {
             u = 1;v = chay;
             while(v-u >1)
             {
                 r = (u+v)/2;
                 if(a[r]<=n)
                     u = r;
                 else v = r;
             }
         }
         v = 1;
         while(v<=u)
         {
             while(a[v]+a[u]<n) v++;
             if(a[v]+a[u]<min)
                 min = a[v]+a[u];
             u--;
         }
         printf("%d\n",min);
    }
    //getch();
}

Code mẫu của khuc_tuan

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <queue>

using namespace std;

#define Rep(i,n) for(int i=0;i<(n);++i)
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Ford(i,a,b) for(int i=(a);i>=(b);--i)
#define fi first
#define se second
#define pb push_back
#define MP make_pair

typedef pair<int,int> PII;
typedef vector<int> VI;

class RepresentableNumbers {
public:
  int getNext(int);
};

int cs[55];
int nc;
int F[55][10][2][2][2];

int go(int pos,int need,int gr,int k1,int k2){
  if(pos == -1) {
    if(need != 0 || (!k1) || (!k2)) return -2;
    else return 0;
  }
  int&ret=F[pos][need][gr][k1][k2];
  if(ret!=-1)return ret;
  // need = need * 10 + cs[pos];
  ret=-2;
  for(int ad=0;ad<10;++ad) if(ad>=cs[pos] || gr){
    int cur = need * 10 + ad;
    for(int x1=k1?1:0;x1<=9;x1+=2){
      for(int x2=k2?1:0;x2<=9;x2+=2){
      int s = x1+x2;    
      if(s == cur || s + 1 == cur){
        int tmp = go(pos-1,cur-s,gr || ad>cs[pos],k1||x1>0,k2||x2>0);
        if(tmp!=-2){
          int z = ad;
          Rep(ii,pos)z*=10;
          if(ret == -2 || ret > z + tmp)
            ret = z + tmp;
        }
      }
      if(x2==0)x2--;
      }
      if(x1==0)x1--;
    }    
  }
  return ret;
}

int RepresentableNumbers::getNext(int X) {
  nc=0;
  while(X>0){
    cs[nc++]=X%10;
    X/=10;
  }
  memset(F,-1,sizeof(F));
  int tmp = go(nc-1,0,0,0,0);
  if(tmp!=-2)return tmp;
  int t=1;
  Rep(i,nc)t*=10;
  return t;
}

RepresentableNumbers run;

int main() {
    while(true){
        string s;
        cin>>s;
        if(s=="[END]")break;
        int n;
        cin>>n;
        cout<< run.getNext(n) << endl;
    }       
}