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 fi='';
      fo='';
      maxn=3500;
      maxp=50;
      base=100000;
      digit=5;
type bignum=array[0..21] of longint;
var n:longint;
    p:array[0..maxp] of longint;
    f:array[0..maxn,0..maxp] of bignum;
    re:array[0..maxn] of bignum;

procedure rf;
begin
     assign(input,fi); reset(input);
     assign(output,fo); rewrite(Output);
end;

procedure prime;
var i,j,t:longint;
begin
     for i:=2 to 200 do
     begin
          t:=0;
          for j:=2 to trunc(sqrt(i)) do
              if i mod j=0 then inc(t);
          if t=0 then
          begin
               inc(p[0]); p[p[0]]:=i;
          end;
     end;
end;

function small(x,y:bignum):boolean;
var i:longint;
begin
     small:=true;
     if x[0]<y[0] then exit;
     if x[0]>y[0] then
     begin
          small:=false; exit;
     end;
     for i:=x[0] downto 1 do
     begin
          if x[i]<y[i] then exit;
          if x[i]>y[i] then
          begin
               small:=false;
               exit;
          end;
     end;
end;

procedure put(var x:bignum;y:bignum);
var i:longint;
begin
     for i:=0 to y[0] do
         x[i]:=y[i];
end;

procedure multi(var x:bignum;y:bignum;z:longint);
var i,mem,max,t:longint;
begin
     mem:=0; max:=y[0];
     for i:=1 to max do
     begin
          t:=y[i]*z+mem;
          x[i]:=t mod base;
          mem:=t div base;
     end;
     if mem>0 then
     begin
          inc(max);
          x[max]:=mem;
     end;
     x[0]:=max;
end;

procedure pr;
var i,j,t:longint; u:bignum;
begin
     prime;
     for i:=0 to maxn do
     begin
         f[i,0,0]:=1; f[i,0,1]:=1;
          re[i,0]:=1;
     end;
     for i:=0 to maxp do
     begin
          f[0,i,0]:=1; f[0,i,1]:=1;
     end;
     for i:=1 to maxn do
         for j:=1 to maxp do
         begin
              if (p[j]>i) or (p[j]=0) then break;
              if small(f[i-1,j],f[i,j-1]) then put(f[i,j],f[i,j-1])
              else put(f[i,j],f[i-1,j]);
              fillchar(t,sizeof(t),0);
              t:=p[j];
              while t<=i do
              begin
                   multi(u,f[i-t,j-1],t);
                   if small(f[i,j],u) then put(f[i,j],u);
                   t:=t*p[j];
              end;
              if small(re[i],f[i,j]) then put(re[i],f[i,j]);
         end;
end;

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

begin
     rf;
     pr;
     wf;
end.

Code mẫu của ladpro98

#include <bits/stdc++.h>

const int N = 3505;
const int BASE = 100000;
const int LENGTH = 5;

using namespace std;

typedef vector<int> bigInt;

void fix(bigInt &a) {
    a.push_back(0);
    for (int i = 0; i + 1 < a.size(); ++i)
        a[i + 1] += a[i] / BASE, a[i] %= BASE;
    while (a.size() > 1 && a.back() == 0) a.pop_back();
}

void operator *= (bigInt &a, int b) {
    for (int i = 0; i < a.size(); ++i) a[i] *= b;
    fix(a);
}

bigInt operator * (bigInt a, int b) {
    a *= b;
    return a;
}

ostream& operator << (ostream& cout, const bigInt &a) {
    cout << a.back();
    for (int i = int(a.size()) - 2; i >= 0; --i)
        cout << setw(LENGTH) << setfill('0') << a[i];
    return cout;
}

bool operator < (bigInt &a, bigInt &b) {
    if (a.size() != b.size()) return a.size() < b.size();
    for (int i = a.size() - 1; i >= 0; --i) if (a[i] != b[i]) return a[i] < b[i];
    return 0;
}

bool isPrime[N];
int P[N];
int nPrime;
bigInt dp[N][N];

void sieve() {
    memset(isPrime, 1, sizeof(isPrime));
    for (int i = 2; i * i < N; ++i) if (isPrime[i]) {
        for (int j = i * i; j < N; j += i)
            isPrime[j] = 0;
    }
    for (int i = 2; i < N; ++i)
        if (isPrime[i]) P[++nPrime] = i;
}

void update(int nxtI, int nxtJ, int i, int j, bigInt &target) {
    if (dp[nxtI][nxtJ] < target)
        dp[nxtI][nxtJ] = target;
}

void preCalc() {
    nPrime = 100;
    for (int i = 0; i < N; ++i) dp[i][0] = bigInt(1, 1);
    for (int i = 1; i <= nPrime; ++i) dp[P[i]][i] = bigInt(1, P[i]);
    bigInt temp;
    for (int i = 0; i < N; ++i) for (int j = 0; j < nPrime && P[j] <= i; ++j) {
        update(i, j + 1, i, j, dp[i][j]);
        for (int p = 1; i + p < N; p *= P[j + 1]) {
            temp = dp[i][j] * p;
            update(i + p, j + 1, i, j, temp);
        }
    }
}

void solve() {
    int n;
    cin >> n;
    bigInt ans (1, 0);
    for (int i = 0; i <= nPrime && P[i] <= n; ++i)
        if (ans < dp[n][i]) ans = dp[n][i];
    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    sieve();
    preCalc();
    int nTest;
    cin >> nTest;
    while (nTest--) solve();
    return 0;
}

Code mẫu của RR

//Written by RR
{$R-,Q-}
{$Mode objfpc}
{$inline on}
uses math;
const
  FINP          =       '';
  FOUT          =       '';
  MAXN          =       3535;
  base          =       100000000;
  lbase         =       8;

type
  big           =       array[0..10] of longint;

operator > (a,b:big) c:boolean;
    var
      i:longint;
    begin
      if a[0]>b[0] then exit(true)
      else if a[0]<b[0] then exit(false);

      for i:=a[0] downto 1 do
        if a[i]>b[i] then exit(true)
        else if a[i]<b[i] then exit(false);

      exit(false);
    end;
operator * (a:big; k:longint) c:big;
    var
      i,nho:longint;
      x:int64;
    begin
      fillchar(c,sizeof(c),0);
      c[0]:=a[0]; nho:=0;
      for i:=1 to c[0] do
        begin
          x:=int64(a[i])*k+nho;
          nho:=x div base;
          c[i]:=x mod base;
        end;
      if nho>0 then
        begin
          inc(c[0]);
          c[c[0]]:=nho;
        end;
    end;

var
  f1,f2         :       text;

procedure print(var a:big);
    var
      c,i,j,x:longint;
    begin
      write(f2,a[a[0]]);
      for i:=a[0]-1 downto 1 do
        begin
          c:=0; x:=a[i];
          while (x>0) do
            begin
              inc(c);
              x:=x div 10;
            end;
          for j:=1 to lbase-c do write(f2,'0');
          write(f2,a[i]);
        end;
      writeln(f2);
    end;

var
  n             :       longint;
  f             :       array[0..MAXN,0..500] of big;
  tmp           :       big;

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

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

const
  MAXV          =       3500;
  MAXP          =       trunc((MAXV/ln(MAXV))*200/100);

var
  dd            :       array[1..MAXV] of byte;
  prime         :       array[1..MAXP] of longint;
  sprime        :       longint;

procedure generate;
    var
      i,j,gh:longint;
    begin
      gh:=trunc(sqrt(MAXV));
      for i:=2 to gh do
      if dd[i]=0 then
        begin
          inc(sprime);
          prime[sprime]:=i;
          j:=i*i;
          while j<=MAXV do
            begin
              dd[j]:=1;
              inc(j,i);
            end;
        end;

      for i:=gh+1 to MAXV do
        if dd[i]=0 then
          begin
            inc(sprime);
            prime[sprime]:=i;
          end;
    end;

procedure init;
    var
      i,k:longint;
      p:int64;
    begin
      for i:=0 to sprime do
        begin
          f[1,i][0]:=1;
          f[1,i][1]:=1;
          f[0,i]:=f[1,i];
        end;
      for i:=2 to 3500 do
        begin
          f[i,0]:=f[1,1];
          for k:=1 to sprime do
            begin
              f[i,k]:=f[i,k-1];
              p:=prime[k];
              while p<=i do
                begin
                  tmp:=f[i-p,k-1]*p;
                  if tmp>f[i,k] then f[i,k]:=tmp;
                  p:=p*prime[k];
                end;
            end;
        end;
    end;

var
  test:longint;

begin
  openF;
  generate;
  init;
  read(f1,test);
  for test:=1 to test do
    begin
      read(f1,n);
      print(f[n,sprime]);
    end;
  closeF;
end.

Code mẫu của hieult

#include <cstdio>
#include <iostream>
#include <cstring>
//#include <conio.h>
#include <cmath>
#define base 1000000000

using namespace std;

struct solon
{
     int so;
     long long a[10];
     solon(){so = 1; a[1] = 0;}
     solon(long long x){so = 1; a[1] = x;}
};

    solon f[3601][505];
    solon Khong = solon(0);
    int u[600],t=1;
    solon Max;


int sosanh(int u,int v){
     if(f[u][v].so > Max.so) return 1;
     if(f[u][v].so < Max.so) return -1;
     for(int i = Max.so;i>=1;i--){
           if(f[u][v].a[i] > Max.a[i]) return 1;
           if(f[u][v].a[i] < Max.a[i]) return -1;
     }
     return 0; 
}

solon tong (solon A,solon B)
{
      int du = 0;
      solon C;
      if(A.so<B.so)
      {
           C = A;
           A = B;
           B = C;
      }
      for(int i = B.so+1;i<=A.so;i++) B.a[i] = 0;
      C.so = A.so;
      for(int i = 1;i<=A.so;i++)
      {

          C.a[i] = (A.a[i]+B.a[i]+du)%base;
          du = (A.a[i]+B.a[i]+du)/base;
      }
      if(du>0) {C.so++; C.a[C.so] = du;}
      return C;
}

solon tichnho(solon A,long long k,int chuso)
{
      solon C;
      long long du = 0;
      C.so = A.so+chuso;
      for(int i = 1;i<=chuso;i++)
           C.a[i] = 0;
      for(int i = chuso+1;i<=chuso+A.so;i++)
      {
           C.a[i] = (A.a[i-chuso]*k+du)%base;
           du = (A.a[i-chuso]*k+du)/base;
      }
      if(du>0) {C.so++; C.a[C.so] = du;}
      return C;
}

solon tich(solon A,solon B)
{
      solon C;
      C.so = 1; C.a[1] = 0;
      for(int i = 1;i<=B.so;i++)
      {
           C = tong(C,tichnho(A,B.a[i],i-1));
      }
      return C;
}

void print(solon A)
{
      printf("%lld",A.a[A.so]);
      for(int i = A.so-1;i>=1;i--)
          printf("%09lld",A.a[i]);
}

solon hieu(solon A,solon B){
      solon C; C.so = A.so;
      for(int i = B.so+1;i<=A.so;i++)
           B.a[i] = 0;
      int du = 0;
      for(int i = 1;i<=A.so;i++){
           C.a[i] = A.a[i]-B.a[i]-du;
           if(C.a[i]<0){
                C.a[i]+=base;
                du = 1;
           }  
      }
      while(C.a[C.so]==0) C.so--;
      if(C.so==0) C.so = 1; 
}

solon chia2(solon A){
      int du = 0;
      long long x;
      for(int i = A.so;i>=1;i--){
           x = A.a[i];
           A.a[i] = (x+du*base)/2;
           du = x%2;
      }
      if(A.a[A.so]==0) A.so--;
      if(A.so==0) A.so++;
      return A;
}


int main()
{

  //  freopen("input.in","r",stdin); 
  //  freopen("output.out","w",stdout);

    u[1] = 2;
    for(int i = 3;i<=3600;i++)
    {
        int flag = 0;
        for(int j = 1;;j++)
        {
            if(u[j]>sqrt(i))
                break;
            else if(i%u[j]==0)
            {
                flag=1;
                break;
            }
        }
        if(flag==0)
        {
            t++;
            u[t]=i;
        }
    }
    //printf("%d",u[t]);
    for(int j = 0;j<=t;j++)
       f[0][j] = solon(1);
    //printf("?\n");   
    for(int i = 1;i<=3500;i++)
    {
        for(int j = 1;;j++)
        {
            if(u[j]>i)
               break;
            else
            {
                Max = solon(0);
                if(sosanh(i - 1, j)>0 )
                    Max = f[i-1][j];
                if(sosanh(i ,j - 1)>0)
                    Max = f[i][j-1];
                int T = u[j];
                while(i-T>=0)
                {
                   // solon X;
                    solon X = tichnho(f[i-T][j-1],T,0);
                    if(X.so > Max.so)
                       Max = X;
                    else if(X.so == Max.so){
                        for(int k = X.so; k >=1; k--) if(X.a[k] > Max.a[k]){
                            Max = X;
                            break;
                            }
                        else if(X.a[k] < Max.a[k]) break;
                    }
                    T=T*u[j];
                }
                f[i][j]=Max;
            }

        }
    }
    //printf("fn\n");
    int test,n;
    scanf("%d",&test);
    for(int ii = 1;ii<=test;ii++)
    {
        scanf("%d",&n);
        if(n==1) printf("1\n");
        else
        {
        Max = solon(0);
        for(int i = 1;i<=t;i++)
        {
           if(sosanh(n ,i)>0)
               Max=f[n][i];
        }
        print(Max);
        printf("\n");
        }
    }
    //getch();
}

Code mẫu của ll931110

{$MODE DELPHI,N+}
program CTNOWN;
const
  input  = '';
  output = '';
  maxn = 3500;
  maxk = 500;
  maxd = 15;
  base = 1000000;
  maxv = 1000000;
var
  F: array[0..maxn,0..maxk] of extended;
  trace: array[0..maxn,0..maxk] of integer;
  d: array[1..maxd] of integer;
  list: array[1..maxk] of integer;
  fi,fo: text;
  n,nlist: integer;
  i,nTest: integer;

procedure openfile;
begin
  assign(fi, input);
    reset(fi);

  assign(fo, output);
    rewrite(fo);
end;

function prime(x: integer): boolean;
var
  i: integer;
begin
  for i := 2 to trunc(sqrt(x)) do if x mod i = 0 then exit(false);
  prime := true;
end;

procedure solve;
var
  i,j,t: integer;
begin
  nlist := 0;
  for i := 2 to maxn do if prime(i) then
    begin
      inc(nlist);
      list[nlist] := i;
    end;

  fillchar(F, sizeof(F), 0);
  for j := 1 to nlist do
    for i := 2 to maxn do
      begin
       F[i,j] := F[i,j - 1];
       trace[i,j] := i;

       t := list[j];
       while i >= t do
         begin
           if F[i,j] < F[i - t,j - 1] + ln(t) then
             begin
               F[i,j] := F[i - t,j - 1] + ln(t);
               trace[i,j] := i - t;
             end;
           t := t * list[j];
         end;
      end;
end;

procedure mul(x: integer);
var
  i: integer;
begin
  if x = 0 then exit;
  for i := 1 to maxd do d[i] := d[i] * x;
  for i := 1 to maxd - 1 do if d[i] >= base then
    begin
      d[i + 1] := d[i + 1] + d[i] div base;
      d[i] := d[i] mod base;
    end;
end;

procedure printresult;
var
  i,u,k,s,tmp: integer;
  st: string;
  max: extended;
begin
  readln(fi, n);
  max := 0;
  u := 0;
  for i := 1 to maxk do
    if max < F[n,i] then
      begin
        max := F[n,i];
        u := i;
      end;

  fillchar(d, sizeof(d), 0);
  d[1] := 1;
  s := n;

  for k := u downto 1 do
    begin
      tmp := trace[s,k];
      mul(s - tmp);
      s := tmp;
    end;

  s := maxd;
  while d[s] = 0 do dec(s);

  write(fo, d[s]);
  for i := s - 1 downto 1 do
    begin
      str(d[i],st);
      for k := 1 to 6 - length(st) do write(fo, 0);
      write(fo, st);
    end;
  writeln(fo);
end;

procedure closefile;
begin
  close(fo);
  close(fi);
end;

begin
  openfile;
  solve;

  readln(fi, nTest);
  for i := 1 to nTest do printresult;

  closefile;
end.

Code mẫu của khuc_tuan

#include <iostream>
#include <cstdio>
using namespace std;


const int COSO = 10000;

struct BigInt {
    int a[20];
    void assign(int x) {
        a[0] = x;
        for(int i=1;i<20;++i)
            a[i] = 0;
    }
    void viet() {
        for(int i=20-1;i>=0;--i) 
            if(a[i] > 0) {
                printf("%d",a[i]);
                for(int j=i-1;j>=0;--j)
                    printf("%04d",a[j]);
                break;
            }
    }
    BigInt opMul(int x){
        BigInt b;
        int nho = 0;
        for(int i=0;i<20;++i) {
            nho += a[i] * x;
            b.a[i] = nho % COSO;
            nho /= COSO;
        }
        return b;
    }
    int opCmp(BigInt b) {
        for(int i=20-1;i>=0;--i) {
            if(a[i] < b.a[i]) return -1;
            if(a[i] > b.a[i]) return 1;
        }   
        return 0;
    }
};

const int MAXN = 3550;

BigInt F[MAXN][300];

int prime[1000];

int main() {

    int nprime = 0;
    for(int i=2;i<300;++i) {
        bool isprime = true;
        for(int j=2;j*j<=i;++j) 
            if(i % j == 0) {
                isprime = false;
                break;
            }
        if(isprime)
            prime[nprime++] = i;
    }   

    for(int i=0;i<MAXN;++i) {
        if(i < 2) F[i][0].assign(1);
        else {
            int z = 2;
            while(2 * z <= i) z *= 2;
            F[i][0].assign(z);
        }
    }

    for(int p=1;p<nprime;++p) {
        for(int i=0;i<MAXN;++i) {
            F[i][p] = F[i][p-1];
            int z = prime[p];
            while(z <= i) {
                BigInt tmp = F[i-z][p-1] .opMul( z );
                if(F[i][p] .opCmp( tmp ) < 0) {
                    F[i][p] = tmp;
                }
                z *= prime[p];
            }
        }
    }

    int ntest, n;
    scanf("%d",&ntest);
    for(int test=0;test<ntest;++test) {
        scanf("%d",&n);
        F[n][nprime-1].viet();
        printf("\n");
    }
    return 0;
}