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 base=10000;
type bignum=array[0..1000] of longint;
var n,m,sum:longint;
    re:bignum; kt:boolean;

procedure rf;
var i,t:longint;
begin
     readln(m,n);
     sum:=0;
     for i:=1 to n do
     begin
          read(t);
          sum:=sum+t;
     end;
    kt:=true;
end;

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

procedure divide(var re:bignum;b:longint);
var mem,i,j:longint;
    t:bignum;
begin
     mem:=0;
     fillchar(t,sizeof(t),0);
     for i:=re[0] downto 1 do
     begin
          j:=re[i]+mem*base;
          if j<b then
          begin
               t[i]:=0;
               mem:=j;
          end
          else
          begin
               t[i]:=j div b;
               mem:=j mod b;
          end;
     end;
     for i:=re[0] downto 1 do
         if t[i]>0 then break;
     re[0]:=i;
     for i:=re[0] downto 1 do
         re[i]:=t[i];
end;

procedure pr;
var i,k,p,j:longint;
begin
     k:=m-(sum+n-1);
     if k<0 then
        begin
        kt:=false;
        exit;
    end;
     p:=n+k;
     if k>n then k:=n;
     re[0]:=1; re[1]:=1;
     for i:=1 to k do
     begin
          multi(re,p-i+1);
          divide(re,i);
     end;
end;

procedure wf;
var i,j,l:longint; s:string;
begin
     if not kt then write(0)
     else
     for i:=re[0] downto 1 do
     begin
          if i<re[0] then
          begin
               str(re[i],s);
               l:=length(s);
               for j:=l+1 to 4 do write(0);
          end;
          write(re[i]);
     end;
end;

begin
     rf;
     pr;
     wf;
end.

Code mẫu của happyboy99x

#include<bits/stdc++.h>
using namespace std;

const int BASE = 10000, MAXLEN = 100;
struct BigInteger {
    int d[MAXLEN], n;

    BigInteger (int x = 0) {
        memset(d, 0, sizeof d); n = 0;
        for(; x != 0; x /= BASE) d[n++] = x % BASE;
    }

    void operator = (int x) {
        memset(d, 0, sizeof d); n = 0;
        for(; x != 0; x /= BASE) d[n++] = x % BASE;
    }

    BigInteger operator + (const BigInteger &x) const {
        BigInteger res; res.n = max(n, x.n);
        int rem = 0;
        for(int i = 0; i < res.n; ++i) {
            int t = x.d[i] + d[i] + rem;
            res.d[i] = t % BASE;
            rem = t / BASE;
        }
        if(rem) res.d[res.n++] = rem;
        return res;
    }
};

ostream& operator << (ostream &out, const BigInteger &x) {
    if(x.n == 0) out << 0;
    else {
        out << x.d[x.n-1];
        for(int i = x.n-2; i >= 0; --i)
            out << setfill('0') << setw(4) << x.d[i];
    }
    return out;
}

const int N = 1000;
BigInteger f[N+1][2];
int n, m, a[N];

int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for(int i = 0; i < m; ++i) cin >> a[i];
    for(int i = 1; i < m; ++i) ++a[i];
    for(int i = 0; i <= n; ++i) f[i][0] = 1;
    for(int j = 1; j <= m; ++j) {
        f[0][j % 2] = 0;
        for(int i = 1; i <= n; ++i) {
            if(i >= a[j-1]) f[i][j % 2] = f[i-1][j % 2] + f[i-a[j-1]][(j-1) % 2];
            else f[i][j % 2] = f[i-1][j % 2];
        }
    }
    cout << f[n][m % 2] << '\n';
    return 0;
}

Code mẫu của ladpro98

program lem6;
uses    math,sysutils;
type    big=string;
const   maxn=1234;
        fi='';
var     sieve:array[1..maxn] of boolean;
        prime,a:array[1..maxn] of longint;
        n,sum,d,m:longint;
        res:big;
procedure input;
var     inp:text;
        i,j,c:longint;
begin
        assign(inp,fi);
        reset(inp);
        readln(inp,n,m);
        for i:=1 to m do
        begin
                read(inp,c);
                inc(sum,c);
        end;
        close(inp);
end;

procedure init;
var     i,j:longint;
begin
        fillchar(sieve,sizeof(sieve),true);
        for i:=2 to maxn do
        begin
                if sieve[i] then
                begin
                        j:=i*i;
                        while j<=maxn do
                        begin
                                sieve[j]:=false;
                                inc(j,i);
                        end;
                end;
        end;
        for i:=2 to maxn do
        if sieve[i] then
        begin
                inc(d);
                prime[d]:=i;
        end;
end;

procedure up(n:longint);
var     i,t:longint;
begin
        for i:=1 to d do
        if prime[i]>n then break
        else
        begin
                t:=prime[i];
                while n>=t do
                begin
                        inc(a[i],n div t);
                        t:=t*prime[i];
                end;
        end;
end;

procedure down(n:longint);
var     i,t:longint;
begin
        for i:=1 to d do
        if prime[i]>n then break
        else
        begin
                t:=prime[i];
                while n>=t do
                begin
                        dec(a[i], n div t);
                        t:=t*prime[i];
                end;
        end;
end;

function mul(a:big; b:longint):big;
var     i,carry,s:longint;
        c:big;
begin
        c:='';
        carry:=0;
        for i:=length(a) downto 1 do
        begin
                s:=(ord(a[i])-48)*b+carry;
                carry:=s div 10;
                c:=chr(s mod 10+48)+c;
        end;
        if carry>0 then
        c:=inttostr(carry)+c;
        exit(c);
end;

function pow(x,i:longint):longint;
var     t:longint;
begin
        if i=1 then exit(x);
        t:=pow(x,i div 2);
        if odd(i) then
                exit(sqr(t)*x);
        exit(sqr(t));

end;

procedure process;
var     k,q,i,p:longint;
begin
        k:=m;
        q:=n-sum+1;
        if k>q then
        begin
                res:='0';
                exit;
        end;
        up(q);
        down(k);
        down(q-k);
        res:='1';
        for i:=1 to d do
        begin
                if a[i]=0 then continue;
                p:=pow(prime[i],a[i]);
                res:=mul(res,p);
        end;
end;

begin
        input;
        init;
        process;
        write(res);
end.

Code mẫu của RR

{$R-,Q-}
uses math;
const
  FINP='';
  FOUT='';
  MAXN=80;
  oo=1000000000;
type
  bigNum=array[0..MAXN] of longint;
var
  a:array[1..1011] of longint;
  d:array[0..1,-10..1011] of bigNum;
  m,n:longint;
  f1,f2:text;
operator +(a:bigNum;var b:bigNum) c:bigNum;
var
  nho,i:longint;
begin
  nho:=0;
  fillchar(c,sizeof(c),0); c[0]:=max(a[0],b[0]);
  for i:=MAXN downto MAXN-c[0]+1 do
    begin
      c[i]:=a[i]+b[i]+nho;
      nho:=c[i] div oo;
      c[i]:=c[i] mod oo;
    end;
  if nho>0 then
    begin
      inc(c[0]);
      c[MAXN-c[0]+1]:=nho;
    end;
end;
procedure print(a:bigNum);
var
  i:longint;
  s:string;
begin
  if a[0]=0 then begin writeln(f2,0); exit; end;
  write(f2,a[MAXN-a[0]+1]);
  for i:=MAXN-a[0]+2 to MAXN do
    begin
      str(a[i],s);
      while length(s)<9 do s:='0'+s;
      write(f2,s);
    end;
  writeln(f2);
end;
procedure openF;
begin
  assign(f1,FINP); reset(f1);
  assign(f2,FOUT); rewrite(f2);
end;
procedure closeF;
begin
  close(f1);
  close(f2);
end;
procedure inp;
var
  i:longint;
begin
  read(f1,n,m);
  for i:=1 to m do read(f1,a[i]);
end;
procedure solve;
var
  i,j:longint;
begin
  for i:=-1 to n do
    begin
      d[0,i,0]:=1;
      d[0,i,MAXN]:=1;
    end;
  for i:=1 to m do
    if i mod 2=1 then
      begin
        fillchar(d[1],sizeof(d[1]),0);
        for j:=1 to n do
        if j>=a[i] then
          d[1,j]:=d[1,j-1]+d[0,j-a[i]-1];
      end
    else {i mod 2=1}
      begin
        fillchar(d[0],sizeof(d[0]),0);
        for j:=1 to n do
        if j>=a[i] then
          d[0,j]:=d[0,j-1]+d[1,j-a[i]-1];
      end;
end;
begin
  openF;
  inp;
  solve;
  print(d[m mod 2,n]);
  closeF;
end.

Code mẫu của hieult

#include<iostream>
#define base 1000000000
#define nm 1001
//#include <conio.h>
using namespace std;
int tren[nm];
int duoi[nm];
long long kq[nm];
int b[nm];
int t=nm-1;
int n,m;
long long power(int x,int y)
{
     if (y==0)return 1;
     long long u=power(x,y/2);
     if (y%2==0)return u*u;
     return u*u*x;
}
void process(int a[],int n)
{
     for (int i=2;i*i<=n;i++)
     {
         if (n%i==0)
            for (int p=1;n%i==0;p++)
            {
                n/=i;
                a[i]++;
            }
     }
     if (n>1)
     a[n]++;
}
void mul(long long k,int&t)
{
     long long temp,nho=0;
     for (int i=1000;i>=t;i--)
     {
         temp=kq[i]*k+nho;
         kq[i]=temp%base;
         nho=temp/base;
     }
     while (nho>0)
     {
           t--;
           kq[t]=nho%base;
           nho=nho/base;
     }
}
int main()
{
    scanf ("%d%d",&n,&m);
    for (int i=0;i<m;i++)
    {
        scanf ("%d",&b[i]);
        n = n-b[i];
    }
    n++;
    if (n<m)
    {
       printf ("0");
       return 0;
    }
    for (int i=n;i>m;i--)
        process(tren,i);
    for (int i=n-m;i>=2;i--)
        process(duoi,i);
    kq[1000]=1;
    for (int i=1000;i>=2;i--)
    {
        tren[i]-=duoi[i];
        if (tren[i])
        {
           long long k=power(i,tren[i]);
           mul(k,t);
        }
    }
    printf ("%lld",kq[t]);
    for (int i=t+1;i<=1000;i++)
        printf("%09lld",kq[i]);
      //  getch();
    //return 0;
}

Code mẫu của ll931110

Program LEM6;
        Const
                input  = '';
                output = '';
        Type
                arr = array[1..200] of byte;
        Var
                p: arr;
                a: array[1..500] of integer;
                F: array[1..500,0..1000] of arr;
                d: array[1..500,0..1000] of integer;
            n,m,t: integer;

Procedure init;
          Var
                fi: text;
                 i: integer;
          Begin
                Assign(fi, input);
                        Reset(fi);
                                Readln(fi, n, m);
                                For i:= 1 to m do read(fi, a[i]);
                Close(fi);
          End;

Procedure addbignum(var x,y,z: arr; m,n: integer);
          Var
                i,k: integer;
          Begin
                If m > n then k:= m else k:= n;
                For i:= 1 to k do
                        Begin
                                z[i]:= z[i] + x[i] + y[i];
                                If z[i] > 99 then
                                        Begin
                                                z[i]:= z[i] - 100;
                                                inc(z[i + 1]);
                                        End;
                        End;
                If z[k + 1] <> 0 then t:= k + 1 else t:= k;
          End;

Procedure optimize;
          Var
                i,k,s: integer;
          Begin
                Fillchar(F[1], sizeof(F[1]), 0);

                Fillchar(p, sizeof(p), 0);

                p[1]:= 1;
                s:= 0;

                F[1,a[1],1]:= 1;
                d[1,a[1]]:= 1;

                For i:= a[1] + 1 to n do
                    Begin
                        addbignum(F[1,i - 1],p,F[1,i],d[1,i - 1],1);
                        d[1,i]:= t;
                    End;

                For i:= 2 to m do
                    Begin
                        s:= s + a[i - 1] + 1;
                        For k:= s to n do
                            Begin
                                  addbignum(F[i,k - 1],F[i - 1,k - a[i] - 1],F[i,k],
                                                        d[i,k - 1],d[i - 1,k - a[i] - 1]);
                                  d[i,k]:= t;
                            End;
                    End;
          End;

Procedure printresult;
          Var
                fo: text;
                 i: integer;
          Begin
                Assign(fo, output);
                        Rewrite(fo);
                        Write(fo, F[m,n,d[m,n]]);
                        For i:= d[m,n] - 1 downto 1 do
                                Begin
                                        if F[m,n,i] < 10 then write(fo, 0);
                                        write(fo, F[m,n,i]);
                                End;
                Close(fo);
          End;

Begin
        init;
        optimize;
        printresult;
End.

Code mẫu của skyvn97

import java.util.*;
import java.lang.*;
import java.math.*;
import java.io.*;
public class Main {
    public static void main(String args[]) {
        InputStream inputStream=System.in;
        OutputStream outputStream=System.out;
        InputReader in=new InputReader(inputStream);
        PrintWriter out=new PrintWriter(outputStream);
        LEM6 solver=new LEM6();
        solver.solve(in,out);
        out.close();
    }
}
class LEM6 {
    public void solve(InputReader in,PrintWriter out) {
        int n=in.nextInt();
        int m=in.nextInt();
        int s=0;
        for (int i=0;i<m;i=i+1) s+=in.nextInt();
        if (s+m-1>n) {
            out.println(0);
            return;
        }
        if (m==1) {
            out.println(n-s+1);
            return;
        }
        BigInteger[][] comb=new BigInteger[m+1][n+1];
        for (int i=0;i<=m;i=i+1)
            for (int j=0;j<=n;j=j+1) {
                if (i>j) comb[i][j]=BigInteger.ZERO;
                if (i==j) comb[i][j]=BigInteger.ONE;
                if (i<j) {
                    if (i==0) comb[i][j]=BigInteger.ONE;
                    else comb[i][j]=comb[i][j-1].add(comb[i-1][j-1]);
                }
            }
        BigInteger res=BigInteger.ZERO;
        for (int x=m-1;x+s<=n;x=x+1) res=res.add(comb[m-2][x-1].multiply(BigInteger.valueOf(n-x-s+1)));
        out.println(res.toString());
    }
}
class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;
    public InputReader(InputStream stream) {
        reader=new BufferedReader(new InputStreamReader(stream),32768);
        tokenizer=null;
    }
    public String next() {
        while (tokenizer==null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer=new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }
    public int nextInt() {
        return Integer.parseInt(next());
    }
    public long nextLong() {
        return Long.parseLong(next());
    }
}

Code mẫu của khuc_tuan

#include <iostream>

using namespace std;

#define coso 100000000
#define maxcs 30

struct BigNum {
    int a[maxcs];
    int cs;
};

int n, m, a[1010];
BigNum F[1010][550];

void cong(BigNum &a, BigNum &b) {
    int cc = a.cs >? b.cs, i, n;
    for(i=n=0;i<cc;++i) a.a[i] += b.a[i] + n, n = a.a[i] / coso, a.a[i] %= coso;
    if(n!=0) a.a[cc++] = n;
    a.cs = cc;
}

int main() {
    cin >> n >> m;
    for(int i=0;i<m;++i) cin >> a[i];
    for(int i=0;i<=n;++i) F[i][0].a[0] = 1, F[i][0].cs = 1;
    for(int i=0;i<=n;++i) for(int j=0;j<m;++j) {
        int ni = i + a[j] + (j==0?0:1);
        if(ni<=n) cong(F[ni][j+1], F[i][j]);
        if(i<n) cong( F[i+1][j], F[i][j]);
    }
    int *res = F[n][m].a, ok = true;
    for(int i=maxcs-1;i>=0;--i) if(res[i]>0) {
        printf("%d", res[i]);
        for(int j=i-1;j>=0;--j) printf("%08d",res[j]);
        ok = false;
        break;
    }
    if(ok) puts("0");
    return 0;
}