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=501;
      maxc=100000000;
var n,m,i,j,re,h,w,t,x,y,k:longint;
    a:array[1..maxn,0..maxn] of longint;

procedure pr;
var i,l,r,min,s,f,hx,sum:longint;
begin
     re:=maxc;
     for i:=1 to n do
         for l:=1 to n-i+1 do
         begin
              if i>re then break;
              r:=l+i-1; min:=maxlongint; hx:=0;
              s:=1; f:=1;
              sum:=a[1,r]-a[1,l-1];
              while (s<=f) and (f<=m) do
              begin
                   if sum<k then
                   begin
                        f:=f+1;
                        sum:=sum+a[f,r]-a[f,l-1];
                   end
                   else
                   begin
                        if f-s+1<min then
                        begin
                             min:=f-s+1;
                             hx:=s;
                        end;
                        sum:=sum-a[s,r]+a[s,l-1];
                s:=s+1;
                   end;
              end;
              if min=maxlongint then continue;
              if (min*i<re) then
              begin
                   re:=min*i;
                   h:=min; w:=i; x:=hx; y:=l;
              end;
         end;
end;

procedure wf;
begin
     assign(output,fo);
     rewrite(output);
     if re<maxc then
     begin
          writeln(re);
          write(x,' ',y,' ',x+h-1,' ',y+w-1);
     end
     else write(-1);
     close(output);
end;

begin
     assign(input,fi);
     reset(input);
     read(m,n,k);
     for i:=1 to m do
         for j:=1 to n do
         begin
              read(t);
              a[i,j]:=a[i,j-1]+t;
         end;
     close(Input);
     pr;
     wf;
end.

Code mẫu của happyboy99x

#include<cstdio>

const int N = 501;
int sum[N][N], a[N][N], m, n, k;

void enter() {
    scanf("%d%d%d", &m, &n, &k);
    for(int i = 1; i <= m; ++i)
        for(int j = 1; j <= n; ++j) {
            scanf("%d", &a[i][j]);
            sum[i][j] = sum[i][j-1] + a[i][j];
        }
}

void solve() {
    int s = m * n + 1, x0, y0, x1, y1;
    for(int l = 1; l <= n; ++l)
        for(int r = l; r <= n; ++r) {
            int now = sum[1][r] - sum[1][l-1];
            for(int t = 1, b = 1; t <= n; ++t) {
                while(b <= n && now < k)
                    ++b, now += sum[b][r] - sum[b][l-1];
                if(now >= k && (r-l+1)*(b-t+1) < s) {
                    s = (r-l+1)*(b-t+1);
                    x0 = t; y0 = l; x1 = b; y1 = r;
                }
                now -= sum[t][r] - sum[t][l-1];
            }
        }
    if(s == m * n + 1) printf("-1\n");
    else printf("%d\n%d %d %d %d\n", s, x0, y0, x1, y1);
}

int main() {
    enter();
    solve();
    return 0;
}

Code mẫu của ladpro98

program helppm;
const   maxn=500;
        fi='';
var     i,j,m,n,k,lim,res,x1,y1,x2,y2,l:longint;
        a,col:array[0..maxn,1..maxn] of longint;
        f,s:array[1..maxn] of longint;
        sum,t,tp:longint;
        inp:text;
begin
        assign(inp,fi);reset(inp);
        readln(inp,m,n,lim);
        for i:=1 to m do
        begin
                for j:=1 to n do read(inp,a[i,j]);
                readln(inp);
        end;
        for i:=1 to m do
        for j:=1 to n do
        col[i,j]:=col[i-1,j]+a[i,j];
        for i:=1 to m do
        begin
                s[i]:=0;
                for j:=1 to n do inc(s[i],a[i,j]);
        end;
        res:=high(longint);
        for i:=1 to m do
        begin
                tp:=0;
                for k:=i to m do
                if (k-i+1<res) then
                begin
                        t:=0;
                        inc(tp,s[k]);
                        if tp<lim then continue;
                        for j:=1 to n do
                        begin
                                f[j]:=col[k,j]-col[i-1,j];
                        end;
                        l:=1;j:=1;sum:=0;
                        for j:=1 to n do
                        begin
                                if sum+f[j]>=lim then break;
                                inc(sum,f[j]);
                        end;
                        while j<=n do
                        begin
                                inc(sum,f[j]);
                                if sum-f[l]>=lim then
                                begin
                                        while sum-f[l]>=lim do
                                        begin
                                                dec(sum,f[l]);
                                                inc(l);
                                        end;
                                end;
                                if sum>=lim then
                                if res>(j-l+1)*(k-i+1) then
                                begin
                                        res:=(j-l+1)*(k-i+1);
                                        x1:=i;x2:=k;
                                        y1:=l;y2:=j;
                                end;
                                inc(j);
                        end;
                end
                else
                break;
        end;
        if res<>high(longint) then
        begin
                writeln(res);
                write(x1,' ',y1,' ',x2,' ',y2);
        end
        else
        write(-1);
end.

Code mẫu của RR

{$R-,Q-}
const
  FINP='';
  FOUT='';
  MAXN=501;
var
  f1,f2:text;
  kq,lu,ld,lr,ll,m,n,k:longint;
  left,down,a:array[0..MAXN,0..MAXN] of longint;
  x:array[1..MAXN] of longint;
procedure openF; inline;
begin
  assign(f1,FINP); reset(f1);
  assign(f2,FOUT); rewrite(f2);
end;
procedure closeF; inline;
begin
  close(f1); close(f2);
end;
procedure inp; inline;
var
  i,j:longint;
begin
  read(f1,m,n,k);
  for i:=1 to m do
  for j:=1 to n do
    read(f1,a[i,j]);
end;
procedure init; inline;
var
  i,j:longint;
begin
  for i:=1 to m do
  for j:=1 to n do
    down[i,j]:=down[i-1,j]+a[i,j];
  for i:=1 to m do
  for j:=1 to n do
    left[i,j]:=left[i,j-1]+a[i,j];
end;
var
  temp:longint;
  height,l,r,i,j,sum:longint;
procedure solve1;
begin
  kq:=maxlongint;
  for l:=1 to m do
  for r:=l to m do
  if r-l<kq then
    begin
      height:=r-l;
      for i:=1 to n do x[i]:=down[r,i]-down[l-1,i];
      j:=1; sum:=x[1];
      if (sum>=k) and (height<kq) then
        begin
          kq:=height;
          lu:=l; ld:=r; ll:=1; lr:=1;
        end;
      for i:=2 to n do
        begin
          inc(sum,x[i]);
          if sum>=k then
            begin
              while sum>=k do
                begin
                  dec(sum,x[j]);
                  inc(j);
                end;
                temp:=(height+1)*(i-j+2);
                if kq>temp then
                begin
                  kq:=temp;
                  lu:=j-1; ld:=i; ll:=l; lr:=r;
                end;
            end;
        end;
    end;
end;
procedure solve2;
begin
  kq:=maxlongint;
  for l:=1 to n do
  for r:=l to n do
  if r-l<kq then
    begin
      height:=r-l;
      for i:=1 to m do x[i]:=left[i,r]-left[i,l-1];
      j:=1; sum:=x[1];
      if (sum>=k) and (height<kq) then
        begin
          kq:=height;
          lu:=l; ld:=r; ll:=1; lr:=1;
        end;
      for i:=2 to m do
        begin
          inc(sum,x[i]);
          if sum>=k then
            begin
              while sum>=k do
                begin
                  dec(sum,x[j]);
                  inc(j);
                end;
              temp:=(height+1)*(i-j+2);
              if kq>temp then
                begin
                  kq:=temp;
                  ll:=j-1; lr:=i; lu:=l; ld:=r;
                end;
            end;
        end;
    end;
end;
procedure ans; inline;
begin
  if kq=maxlongint then begin writeln(f2,-1); exit; end;
  writeln(f2,kq);
  writeln(f2,ll,' ',lu,' ',lr,' ',ld);
end;
begin
  openF;
  inp;
  init;
  if m<n then solve1
  else solve2;
  ans;
  closeF;
end.

Code mẫu của hieult

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

int main()
{
    //freopen("1.in","r",stdin);
    int n,m,area,a[502][502],x,x1,y1,x2,y2;
    long long are = 0;
    scanf("%d %d %d",&n,&m,&area);
    for(int i = 0;i<m;i++)
        a[0][i] = 0;
    for(int i = 1;i<=n;i++)
    {
        a[i][0] = 0;
        for(int j = 1;j<=m;j++)
        {
            scanf("%d",&x);
            a[i][j] = a[i][j-1]+x;
            are = are + x;
           // printf("%d ",a[i][j]);
        }
       // printf("\n");
    }
    if(are<area)
         printf("-1");
    else
    {
    x1 = 1;y1=1;x2 = m;y2 = n; int min = n*m;
    for(int i = 1;i<=n;i++)
        for(int j = i;j<=n;j++)
        {
            int u = 1 , v = 1;long long tong = 0;
            while(v<=n)
            {
                tong  = tong + a[v][j] - a[v][i-1];
                if(tong >=area)
                {
                    if((v-u+1) * (j-i+1)<min)
                    {
                        min = (v-u+1) * (j-i+1);
                        x1  = i; y1 = u; x2 = j; y2 = v;
                    }
                    u++;
                    while(u<=v)
                    {
                        tong = tong - a[u-1][j]+a[u-1][i-1];
                        if(tong >=area)
                        {
                            if((v-u+1) * (j-i+1)<min)
                            {
                                 min = (v-u+1) * (j-i+1);
                                 x1  = i; y1 = u; x2 = j; y2 = v;
                            }
                        }
                        else break;
                        u++;
                    }
                }
               // printf("%d %d\n",u,v);
                v++;
            }
        }
    printf("%d\n%d %d %d %d",min,y1,x1,y2,x2);
    }
  // getch();
}

Code mẫu của ll931110

{$MODE DELPHI}
{$inline on}
Program HELPPM;
Const
  input  = '';
  output = '';
  maxn = 503;
  maxv = 1000000;
Var
  a: array[0..maxn,0..maxn] of integer;
  m,n,k: integer;
  area,lx,ly,rx,ry: integer;

Procedure init;inline;
Var
  f: text;
  i,j: integer;
Begin
  Assign(f, input);
    Reset(f);

  Readln(f, m, n, k);
  For i:= 1 to m do
    For j:= 1 to n do read(f, a[i,j]);

  Close(f);

  For i:= 1 to m do
    For j:= 2 to n do a[i,j]:= a[i,j] + a[i,j - 1];

  For j:= 1 to n do
    For i:= 2 to m do a[i,j]:= a[i,j] + a[i - 1,j];
End;

Procedure solve;inline;
Var
  i,j,inf,sup: integer;
Begin
  area:= maxv;
  For i:= 1 to n do
    For j:= i to n do
      If area > j - i + 1 then
        Begin
          inf:= 1;
          sup:= 1;

          Repeat
            While (a[sup,j] - a[sup,i - 1] - a[inf - 1,j] + a[inf - 1,i - 1] < k) and (sup <= m) do inc(sup);
            If sup > m then break;

            While a[sup,j] - a[sup,i - 1] - a[inf,j] + a[inf,i - 1] >= k do inc(inf);

            If area > (j - i + 1) * (sup - inf + 1) then
              Begin
                area:= (j - i + 1) * (sup - inf + 1);
                lx:= inf;
                ly:= i;
                rx:= sup;
                ry:= j;
              End;

            inc(sup);
          Until sup > m;
        End;
End;

Procedure printresult;inline;
Var
  f: text;
Begin
  Assign(f, output);
    Rewrite(f);

  If area = maxv then writeln(f, -1)
  else
    Begin
      Writeln(f, area);
      Writeln(f, lx, ' ', ly, ' ', rx, ' ', ry);
    End;

  Close(f);
End;

Begin
  init;
  solve;
  printresult;
End.

Code mẫu của skyvn97

#include<cstdio>
    #define MAX 505
    int a[MAX][MAX];
    int s[MAX][MAX];
    int k;
    int m,n;
    int res;
    int tx,ty,bx,by;
    void init(void) {
    scanf("%d",&m);
    scanf("%d",&n);
    scanf("%d",&k);
    int i,j;
    for (i=0;i<=m;i=i+1) s[i][0]=0;
    for (i=0;i<=n;i=i+1) s[0][i]=0;
    for (i=1;i<=m;i=i+1)
    for (j=1;j<=n;j=j+1) {
    scanf("%d",&a[i][j]);
    s[i][j]=s[i][j-1]+s[i-1][j]-s[i-1][j-1]+a[i][j];
    }
    res=m*n+1;
    }
    void count(const int &b,const int &c) {
    int i,j;
    i=1;j=1;
    if (res<=c-b+1) return;
    if (s[c][n]-s[b-1][n]<k) return;
    while (true) {
    if (s[c][j]-s[c][i-1]-s[b-1][j]+s[b-1][i-1]>=k) {
    if (res>(c-b+1)*(j-i+1)) {
    res=(c-b+1)*(j-i+1);
    tx=b;ty=i;bx=c;by=j;
    }
    i++;
    }
    else {
    if (j>=n) break;
    else j++;
    }
    }
    }
    void process(void) {
    int i,j;
    for (i=1;i<=m;i=i+1)
    for (j=i;j<=m;j=j+1)
    count(i,j);
    if (res>m*n) printf("-1");
    else printf("%d\n%d %d %d %d",res,tx,ty,bx,by);
    }
    int main(void) {
    init();
    process();
    return 0;
    }

Code mẫu của khuc_tuan

#include <stdio.h>
#include <string.h>

int m, n, k;
int a[505][505];
long long d[505];

int main() {
    scanf("%d%d%d", &m, &n, &k);
    for(int i=0;i<m;++i)
        for(int j=0;j<n;++j)
            scanf("%d", a[i]+j);
    int smin = 1000000000, x, y, u, v;
    int inf = 1000000000;
    for(int i1=0;i1<m;++i1) {
        memset( d, 0, sizeof(d));
        for(int i2=i1;i2<m;++i2) {
            int mm = 1000000000;
            int bd, kt;
            long long cur = 0;
            for(int j=0, tr=0;j<n;++j) {
                d[j] += a[i2][j];
                cur += d[j];
                while(cur-d[tr]>=k) { cur -= d[tr]; ++tr; }
                if(cur>=k && mm>j-tr) {
                    mm = j - tr;
                    bd = tr;
                    kt = j;
                }
            }
            ++mm;
            if(mm<1000000000 && mm * (i2-i1+1) < smin) {
                smin = mm * (i2-i1+1);
                x = i1;
                u = i2;
                y = bd;
                v = kt;
            }
        }
    }
    if(smin==1000000000) printf("-1\n");
    else printf("%d\n%d %d %d %d", smin, x+1, y+1, u+1, v+1);
    return 0;
}