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>
using namespace std;

long long P[222];

void decode(int a[],int n,long long k)
{
    for (int i=0;i<n;i++)
        for (a[i]=0;k>P[n-1-i];a[i]++)
            k-=P[n-1-i];
}

long long countPalin(int n,long long lim)
{
    if (n==1) return 0;
    int a[222],half=(n+1)/2,cmp=0;
    long long res=0;
    decode(a,n,lim);
    for (int i=0;i<half;i++)
        if (a[i]) res+=P[half-1-i]*a[i];
    for (int i=half-1;i>=0 && !cmp;i--)
        if (a[i]>a[n-1-i]) cmp=1;
        else
            if (a[i]<a[n-1-i]) cmp=-1;
    if (cmp<=0) res++;
    return res; 
}

long long findNewK(int n,long long k)
{
    long long last=0,res=k;
    while (1)
    {
        long long x=countPalin(n,res);
        if (res-x>=k) break;
        res+=x-last;
        last=x;
    }
    return res;
}

void solve(int a[],int n,int p,long long k)
{
    if (p) 
    {
        solve(a,(n+1)/2,p-1,k);
        for (int i=0;i<n-1-i;i++) a[n-1-i]=a[i];
        return;
    }
    k=findNewK(n,k);
    decode(a,n,k);
}

int main()
{
    int n,p,a[222];
    long long k;
    cin >> n >> p >> k;
    for (int i=0;i<=n;i++) P[i]=(i?(i>9?P[9]:P[i-1]*26):1);
    solve(a,n,p,k);
    for (int i=0;i<n;i++) cout << char(a[i]+97);
    cout << endl;
}

Code mẫu của ladpro98

#include <bits/stdc++.h>

using namespace std;

const int N = 222;
const long long INF = (long long)1e15;

long long power[N];

string solve(int n, int k) {
    power[0] = 1;
    for (int i = 1; i < n; ++i) power[i] = min(power[i - 1] * 26, INF);
    string s (n, char('a' + k - 1));
    if (n == 1) return s;
    bool canBePalin = 1;
    for (int i = 0; i < n; ++i) {
        long long all = power[n - i - 1];
        for (char c = 'a'; c <= 'z'; ++c) {
            long long palin = 0;
            if (i <= (n - 1) / 2)
                palin = power[(n - 1) / 2 - i];
            else if (canBePalin && c == s[n - i - 1])
                palin = 1;
            long long nonPalin = all;
            if (all != INF) nonPalin -= palin;
            if (k > nonPalin) {
                k -= nonPalin;
            } else {
                s[i] = c;
                if (i > (n - 1) / 2)
                    canBePalin &= c == s[n - i - 1];
                break;
            }
        }
    }
    return s;
}

int main() {
    int n, deg, k;
    cin >> n >> deg >> k;
    int m = n;
    vector<int> save;
    for (int i = 0; i < deg; ++i) {
        save.push_back(m & 1);
        m = (m + 1) / 2;
    }
    string s = solve(m, k);
    for (int i = 0; i < deg; ++i) {
        string t = s;
        reverse(t.begin(), t.end());
        if (save.back())
            s += t.substr(1);
        else
            s += t;
        save.pop_back();
    }
    cout << s << endl;
    return 0;
}

Code mẫu của RR

//Written by Nguyen Thanh Trung
{$R+,Q+}
{$Mode objFPC}
uses math;
const
  FINP          =       '';
  FOUT          =       '';
  MAXN          =       222;
  oo            =       1000000001;
var
  f1,f2         :       text;
  n,p,k         :       longint;
  s             :       string;
  lt26          :       array[0..MAXN] of int64;

procedure openF;
    begin
      assign(f1,FINP); reset(f1);
      assign(f2,FOUT); rewrite(f2);
    end;
procedure closeF;
    begin
      close(f1);
      close(f2);
    end;
procedure build1(l,kk:longint);
    var
      i:longint;
      now:int64;
      c:char;
    begin
      s:='';
      now:=0;
      for i:=1 to l do
        begin
          c:='a';
          while now+lt26[l-i]<=kk do
            begin
              c:=succ(c);
              inc(now,lt26[l-i]);
            end;
          s:=s+c;
        end;
    end;
function equal(a,b:char):longint;
    begin
      if a=b then exit(1) else exit(0);
    end;
procedure build2(l,kk,start:longint;b:boolean);
    var
      i,j:longint;
      now:int64;
      c:char;
    begin
      j:=start;
      now:=0; dec(kk);
      for i:=1 to l do
      begin
        if b then
          begin
            c:='a';
            while now+lt26[l-i]-equal(c,s[j])<=kk do
              begin
                inc(now,lt26[l-i]-equal(c,s[j]));
                c:=succ(c);
              end;
            s:=s+c;
            b:=b and (c=s[j]);
          end
        else //if not b
          begin
            c:='a';
            while now+lt26[l-i]<=kk do
              begin
                c:=succ(c);
                inc(now,lt26[l-i]);
              end;
            s:=s+c;
          end;
        dec(j);
      end;
    end;
procedure build3(n,p,l:longint);
    var
      i,j:longint;
    begin
      if n=l then exit;
      build3((n+1)>>1,p,l);
      if n and 1=1 then
        begin
          j:=length(s)-1;
          for i:=n>>1 downto 1 do
            begin
              s:=s+s[j];
              dec(j);
            end;
        end
      else //n and 1=0
        begin
          j:=length(s);
          for i:=n>>1 downto 1 do
            begin
              s:=s+s[j];
              dec(j);
            end;
        end;
    end;
procedure solve;
    var
      l,i:longint;
      x:int64;
    begin
      l:=n;
      for p:=1 to p do
        l:=(l+1)>>1;
      if l=1 then s:=char(ord('a')+k-1)
      else
        begin
          x:=1;
          for i:=l>>1 downto 1 do begin x*=26; if x>oo then x:=oo; end;
          x-=1;
          build1((l+1)>>1,(k-1) div x);
          k:=k mod x; if k=0 then k:=x;
        end;
      build2(l-(l+1)>>1,k,l>>1,true);
      build3(n,p,l);
      writeln(f2,s);
    end;
procedure init;
    var
      i:longint;
    begin
      lt26[0]:=1;
      for i:=1 to MAXN do
        begin
          lt26[i]:=lt26[i-1]*26;
          if lt26[i]>oo then lt26[i]:=oo;
        end;
    end;

begin
  openF;
  read(f1,n,p,k);
  if n=1 then
    begin
      writeln(char(ord('a')+k-1));
      closeF; halt;
    end;
  init;
  solve;
  closeF;
end.

Code mẫu của ll931110

#include <algorithm>
#include <bitset>
#include <cstring>
#include <deque>
#include <fstream>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <vector>
using namespace std;

int n,k,num,len;
int q[202],s[202];
int f[12];
char list[202];

int get(int a,int m)
{
    if (a == 1) return (m - 1);   
    int prefix = (a + 1)/2; 
    int inf = 0,sup = 2000000000,fin = -1;
    while (inf <= sup)
    {
          int med = (inf + sup)/2;
          int tmp = med;
          for (int i = 0; i < a - prefix; i++) tmp /= 26;
          int palin = tmp - 1;          

          tmp = med;
          for (int i = a; i > 0; i--)
          {
              s[i] = tmp % 26; tmp /= 26;
          };
          for (int i = 1; i <= prefix; i++) q[i] = s[i];
          for (int i = prefix + 1; i <= a; i++) q[i] = q[a + 1 - i];
          bool lower = true;
          for (int i = 1; i <= a; i++) if (q[i] != s[i])
          {
              if (q[i] > s[i]) lower = false;
              break;
          };

          if (lower) palin++;
          if (med - palin < num) inf = med + 1; else
          {
             fin = med;  sup = med - 1;
          };
    };
//    cout << fin;
    return fin;
};

int main()
{
//  freopen("lsp.in","r",stdin);
//  freopen("lsp.ou","w",stdout);
    scanf("%d %d %d", &n, &k, &num);

    int len = n;
    for (int i = k; i > 0; i--) len = (len + 1)/2;
    int ret = get(len,num);

    for (int i = len; i > 0; i--)
    {
        q[i] = ret % 26;  ret /= 26;        
    };
    f[k] = n;
    for (int i = k - 1; i > 0; i--) f[i] = (f[i + 1] + 1)/2;

    for (int i = 1; i <= k; i++)
      for (int j = (f[i] + 1)/2 + 1; j <= f[i]; j++) q[j] = q[f[i] + 1 - j];

    for (int i = 1; i <= n; i++) list[i] = q[i] + 'a';
    for (int i = 1; i <= n; i++) printf("%c", list[i]);
};

Code mẫu của skyvn97

#include<cstdio>
#include<iostream>
#define MAX   222
#define FOR(i,a,b) for (int i=(a);i<=(b);i=i+1)
#define FORD(i,b,a) for (int i=(b);i>=(a);i=i-1)
#define REP(i,n) for (int i=0;i<(n);i=i+1)
using namespace std;
typedef long long ll;
const ll INF=(ll)1e9+7LL;
ll pw[MAX];
int len[MAX];
int n,p,l;
char s[MAX];
ll k;
void init(void) {
    cin >> n >> p >> k;
    pw[0]=1LL;
    FOR(i,1,n) pw[i]=min(pw[i-1]*26,INF);   
    len[0]=n;
    FOR(i,1,p) len[i]=(len[i-1]+1)/2;
    l=len[p];
}
void buildstr(void) {
    if (l==1) {
        s[1]=k-1;
        return;
    }
    //printf("Build %d\n",l);
    bool canpalin=true;
    ll nles,ples;
    FOR(i,1,l) {
        //printf("Step %d\n",i);
        REP(j,27) { 
            if (i<=(l+1)/2) nles=min(j*pw[(l+1)/2-i],INF)*(pw[l-(l+1)/2]-1);
            else nles=j*pw[l-i]-(canpalin && j>s[l+1-i]);
            //printf("\tCase %d has %lld les\n",j,nles);
            if (nles>=k) {
                k-=ples;
                s[i]=j-1;
                if (i>(l+1)/2) canpalin=(canpalin && s[i]==s[l+1-i]);
                break;
            }
            else ples=nles;
        }       
    }
}
void buildall(void) {
    int cur=l;
    FORD(i,p-1,0) {
        FOR(j,cur+1,len[i]) s[j]=s[len[i]-j+1];
        cur=len[i];
    }
    FOR(i,1,n) printf("%c",s[i]+'a');   
}
int main(void) {
    ios::sync_with_stdio(false);
    init();
    buildstr();
    buildall();
    return 0;
}