Đề bài: http://vn.spoj.com/problems/VMDAOBIT/

Thuật toán:

• (đang cập nhập)

Code:

const   fi='';
        fo='';
        maxn=100;
type    point=record
                x,y:integer;
                end;
var     m,n,i,j,dem:longint;
        a,b:array[1..maxn,1..maxn] of byte;
        res:array[1..maxn*maxn] of point;
        kiemtra:boolean;
procedure update(x,y:integer);
var     i,j:integer;
begin
        for i:=x to x+2 do
                for j:=y to y+2 do
                        if b[i,j]=1 then b[i,j]:=0 else b[i,j]:=1;
end;
procedure xaydung;
var     i,j:integer;
begin
  for i:=1 to m do
  begin
    for j:=1 to n-2 do
        begin
            if b[i,j]=1 then
                begin
                    inc(dem);
                    res[dem].x:=i;
                    res[dem].y:=j;
                    update(i,j);
                end;
        end;
    for j:=n-1 to n do
        if b[i,j]=1 then begin kiemtra:=false; exit; end;
  end;
end;
procedure init;
begin
    b:=a;
    dem:=0;
    kiemtra:=true;
end;
begin
    assign(input,fi);reset(input);
    assigN(output,fo);rewrite(output);

    readln(m,n);
    for i:=1 to m do
        for j:=1 to n do read(a[i,j]);

    init;
    xaydung;

    if kiemtra then
        begin
                writeln(dem);
                for i:=1 to dem do writeln(res[i].x,' ',res[i].y);
        end else write(-1);

    close(input);close(output);
end.

Đề bài: http://vn.spoj.com/problems/CRYPTKEY/

Thuật toán:

• (đang cập nhập)

Code

const
  fi='';
  fo='';
  maxn=50003;
var
  a : array[1..maxn] of int64;
  b : array[1..maxn] of int64;
  c : array[1..100000] of int64;
  uoc : array[1..100000] of int64;
  sl : array[1..100000] of longint;
  i,j,n,t,tt : longint;
  duoc : longint;
  kt : boolean;
  res,k : int64;
procedure enter;
begin
  assign(input,fi);reset(input);
  assign(output,fo);rewrite(output);
end;
procedure pt(x : int64);
  var i,j,dem : longint;
  begin
    for i:=2 to trunc(sqrt(x)) do
      if x mod i = 0 then
        begin
          dem := 0;
          while x mod i = 0 do
            begin
              inc(dem);
              x := x div i;
            end;
          duoc := duoc +1;
          uoc[duoc] := i;
          sl[duoc] := dem;
        end;
    if x>1 then
      begin
        inc(duoc);
        uoc[duoc] := x;
        sl[duoc] := 1;
      end;
  end;
function power(x,y:longint):int64;
  var i : longint;
  begin
    power:=1;
    for i:=1 to y do power := power * x;
  end;
function gcd(a,b : int64) : int64;
  var r : int64;
  begin
      while b<>0 do
        begin
          r := a mod b;
          a:=b;
          b:=r;
        end;
      exit(a);
  end;
function lcm(a,b : int64) : int64;
  var x,y : int64;
  begin
    {x := a; y:=b;}
    lcm := (a div gcd(a,b)) *b;
  end;
procedure sol;
var i,j : longint;
    tg : int64;
    dem : longint;
begin
  read(t);
  for tt :=1 to t do
    begin
      fillchar(a,sizeof(a),0);
      fillchar(uoc,sizeof(uoc),0);
      fillchar(sl,sizeof(sl),0);
      kt := true;
      read(n);
      for i:=1 to n do read(a[i]);
      readln(k);
      pt(k);
      for i:=1 to duoc do
      begin
        tg := power(uoc[i],sl[i]);
        dem := 0;
        for j:=1 to n do
          if a[j] mod tg = 0 then
           begin
             inc(dem); b[dem] := a[j];
           end;
        if dem=0 then begin kt := false; break; end;
        tg := b[1];
        for j:=2 to dem do
          tg := gcd(tg,b[j]);
        c[i] := tg;
      end;
      if not kt then begin writeln('NO'); continue; end;
      res := c[1];
      for i:=2 to duoc do res := lcm(res,c[i]);
      if res=k then
      writeln('YES') else writeln('NO');
    end;
  close(input);close(output);
end;
begin
  enter;
  sol;
end.

Đề bài: http://vn.spoj.com/problems/DHTABLE2/

Thuật toán:

• (đang cập nhập)

Code:

var t:int64;
    n:integer;
    f:text;
procedure nhap;
begin
   assign(f,'');reset(f);
   read(f,n,t);
   close(f);
end;

procedure xuli;
var i:longint;
    x,l:string;
    mu,k,m
    :int64;
begin
    assign(f,'');rewrite(f);
    mu:=1; i:=0;
    while t>9*mu*(i+1) do
          begin
              dec(t,9*mu*(i+1));
              inc(i);
              mu:=mu*10;
          end;
    inc(i);
    k:=mu+(t-1) div i;
    t:=t mod i;
    x:='';
    while (length(x)0) do
        begin
            m:=k;
            str(m,l);
            x:=l+x;
            if t>0 then
               begin
                   delete(x,t+1,i-t);
                   t:=0;
               end;
            dec(k);
        end;
    while length(x)>n do delete(x,1,1);
    while length(x)