Почему в примере ответ 2 ? Непонимаю условия!

My answer is
n=2 : 140
n=3 : 1400
n=4 : 14000


Edited by author 23.12.2005 20:41

поменял long long на int64 все прошло, полный Бред!

was doing a silly spelling mistake.

Edited by author 06.08.2011 22:32

Edited by author 06.08.2011 22:32

In 0.140 sec.

In fact, you should consider, that until the last collision, fireball can fly through the wizard and the target.

Sorry once again...
   Problem Author

На самом деле надо считать, что вплоть до последнего столкновения файрболл может пролетать сквозь чародея и цель.

Еще раз прошу прощения.
   Автор задачи.

Try this: 09999999999000

WA7, and I want to know is it round problem or fault in my data structure.
Tests like test7 will be very helpful for me


Edited by author 06.08.2011 14:27

const
  zs=699967;
type
  size=record
    di,zhuan:longint;
    x,y:integer;
  end;
var
  dz:array[0..14329,1..2] of longint;
  dui:Array[0..14329,1..2] of int64;
  hash:array[0..700000] of size;
  xx,s1,s2,pd,n,m,i1,i2,zhuan,l1,l2,nzhuan,k:longint;
  d:Array[1..2] of longint;
  map:Array[0..13,0..13] of  char;
  san:array[0..13] of longint;
  zhi:int64;
  i,j,ii:longint;
procedure swap(var a,b:longint);
var
  t:longint;
begin
  t:=a; a:=b; b:=t;
end;
function zb(nzhuan:longint):longint;
begin
  zb:=nzhuan mod  zs;
  while (hash[zb].x=i) and (hash[zb].y=j) and (hash[zb].zhuan<>nzhuan) do
  begin
    inc(zb);
    zb:=zb mod zs;
  end;
end;


procedure jiaru(nzhuan:longint;zhi:int64);
var
  nzb:longint;
begin
  if ((map[i+1,j]='*')and (nzhuan div san[j-1] mod 3>0))
    or ((map[i,j+1]='*')and (nzhuan div san[j] mod 3>0)) then
    exit;
  nzb:=zb(nzhuan);
  if (hash[nzb].x<>i)or (hash[nzb].y<>j) then
  begin
    inc(d[i2]);
    dz[d[i2],i2]:=nzhuan;
    dui[d[i2],i2]:=zhi;
    hash[nzb].x:=i; hash[nzb].y:=j;
    hash[nzb].di:=d[i2];
    hash[nzb].zhuan:=nzhuan;
  end else begin
    inc(dui[hash[nzb].di,i2],zhi);
  end;
end;
begin
  san[0]:=1;
  for i:=1 to 13 do san[i]:=san[i-1]*3;
  readln(n,m);
  for i:=1 to n do
  begin
    for j:=1 to m do
    begin
      read(map[i,j]);
      if i=1 then map[i-1,j]:='*';
      if i=n then map[i+1,j]:='*';
      if j=1 then map[i,j-1]:='*';
      if j=m then map[i,j+1]:='*';
    end;
    readln;
  end;
  map[0,0]:='*'; map[0,m]:='*';
  for i:=0 to 600000 do
  hash[i].x:=0;
  dui[1,1]:=1; dz[1,1]:=0;
  i1:=1; i2:=2;
  d[1]:=1;
  for i:=1 to n do
  begin
    for j:=1 to m do
    begin
      d[i2]:=0;
      for ii:=1 to d[i1] do
      begin
        zhuan:=dz[ii,i1];
        zhi:=dui[ii,i1];
        l1:=zhuan div san[j-1] mod 3;
        l2:=zhuan div san[j] mod 3;
        if l1+l2=3 then
        begin
          if (l1=1)and (l2=2)and ((i<>n) or (j<>m)) then continue;
          nzhuan:=zhuan-san[j-1]*l1-san[j]*l2;
          jiaru(nzhuan,zhi);
        end else
        if l1+l2=0 then
        begin
          if map[i,j]='.' then
          nzhuan:=zhuan+san[j-1]*1+san[j]*2 else
          nzhuan:=zhuan;

          jiaru(nzhuan,zhi);
        end else if l1=l2 then
        begin
          if l1=1 then
          begin
            pd:=0;
            k:=j+1;
            while pd<>1 do
            begin
              inc(k);
              xx:=zhuan div san[k-1] mod 3;
              if xx=1 then dec(pd);
              if xx=2 then inc(pd);
              if pd=1 then s1:=k;
            end;
            nzhuan:=zhuan-san[j-1]-san[j]-san[s1-1];
            jiaru(nzhuan,zhi);
          end else begin
            pd:=0;
            k:=j;
            while pd<>1 do
            begin
              dec(k);
              xx:=zhuan div san[k-1] mod 3 ;
              if xx=1 then inc(pd);
              if xx=2 then dec(pd);
              if pd=1 then s2:=k;
            end;
            nzhuan:=zhuan-san[j-1]*2-san[j]*2+san[s2-1];
            jiaru(nzhuan,zhi);
          end;
        end else
        begin
          jiaru(zhuan,zhi);
          nzhuan:=zhuan-san[j-1]*l1-san[j]*l2;
          swap(l1,l2);
          nzhuan:=nzhuan+san[j-1]*l1+san[j]*l2;
          jiaru(nzhuan,zhi);
        end;
      end;
      swap(i1,i2);
    end;
    j:=m;
    d[i2]:=0;
    for ii:=1 to d[i1] do
    begin
      if (dz[ii,i1]div san[j]) >0then
      begin
        writeln('nimeia');
        continue;
      end;
      inc(d[i2]);
      dz[d[i2],i2]:=dz[ii,i1]*3;
      dui[d[i2],i2]:=dui[ii,i1];

    end;
    swap(i1,i2);
  end;
  if d[i1]<>1 then writeln('0')else
  writeln(dui[1,i1]);
end.

Where can be a problem?
Wrong answer 3
Pascal

program coun;

var a:array[1..1000000] of extended;
i,j:integer; k:extended;
begin
i:=0;
repeat
 read(k);
 i:=i+1;
 a[i]:=sqrt(k);
until eof();
for j:=i-1 downto 1 do
begin
writeln(a[j]:0:5);
end;
end.

why is solution with long double not working, long double must have 18 digits according to ISO c++, which must be enough to solve this problem

i've calculated everything using g++ and long double, and formed a constant and got AC, which means that system compiler is bad..

what s test 3 ?

#include <iostream>
#include <cmath>
#include <iomanip>

using namespace std;

int main(){

    long double x;
    int i=0;
    long double tab[32768];

    cout << setprecision(4) << fixed;
    cin >> x;
    while(!cin.eof()){
              tab[i]=sqrt(x);
              i+=1;
              cin >> x;
              }
    for(int k=0;k<i;k++){
                   cout << tab[i-1-k] << endl;
                   }
                   getchar();getchar();
    }

what s test 10 ? here s my code, if U know what s wrong with the code without test 10 i would be most greatful to know it.
thx - tom

and btw what s the answer for n = 0 ?
1 or 0 ?
anyway i checked it and it s not test 10.

#include<iostream>
#include<cstring>
#include<string>

using namespace std;

int main(){

    int k,n;
    string a;
    cin >> n;
    cin >> a;
    int i;
    i=0; k=0;
    int x =1;
    while(a[i]=='!'){
            i++;
            k++;
            }
    if(!(n%k==0)){
            if(n%k==1){
                 while(n!=1){
                       x*=n;
                       n-=i;
                       }
                 }
            else x=n;
            }
    else{
          while(n!=0){
               x*=n;
               n-=i;
               }
         }
    cout << x;
    getchar();getchar();
}



Edited by author 05.08.2011 14:34

Input:- 10 20 3 6
Output:- 140

Input:- 30 65 45 12
Output:- 5310

Input:- 99 99 2 100
Output:- 29007

Input:- 50 50 1 87
Output:- 12850

Input:- 33 66 43 100
Output:- 9372

Input:- 54 100 2 77
Output:- 18996

Input:- 1 1 100 54
Output:- 139

Edited by author 30.04.2009 12:11

Edited by author 30.04.2009 12:11

i dont get it at all...it looks good though. at least to me
help pls

#include<iostream>

using namespace std;

int main(){

    int m,n,a;
    cin >> m >> n;
    a=(m*n)%2;
    if(a==0) cout << "[:=[first]";
    else if(a==1) cout << "[second]:=]";
}

how we get 19 in example? don't understand how got odd number.
my answer is 20.

Can you help me with this, I'm getting this error on 12th test data.

Thanks