Is there any way that we can download all our source codes in one big archive, without entering the ID a hundred times?

This problem is so easy ^^!

Can anybody give some stress test for this task?pls

AC code

***
var
a : array[1..500001] of integer;
function BinarySearch(l, r, q : integer): boolean;
begin
if (r - l = 1) then
if ((a[l] = q) or (a[r] = q)) then
result := true
else
result := false
else
if a[(l+r) div 2] > q then
result := BinarySearch(l, (l+r) div 2, q)
else
result := BinarySearch((l+r) div 2, r, q);
end;
var
n, i, k, x : integer;
begin
readln(n);
for i := 1 to n do
read(a[i]);
readln(k);
for i := 1 to k do
begin
read(x);
if BinarySearch(1, n, 10000-x) then
begin
writeln('YES');
exit;
end
end;
writeln('NO');
end.
***

Why do I have access violation on test 3?
There is my cod:

var t,i,j,l:longint;
a,b,s,x:int64;
x1,x2,r:extended;
m,d:array [1..124776567] of int64;
k:array [1..20] of int64;
smena:boolean;

begin
 readln(t);
 for i:=1 to t do
  begin
   readln(a,b);
   if b mod a<>0 then k[i]:=0
    else if a=b then k[i]:=1
    else begin
     x:=round(b/a);
     s:=x;
     l:=0;
     smena:=true;
     j:=1;
     x1:=1;
     x2:=1;
     while (j<100) or (x2-x1>1e-5) do begin
      r:=x2;
      x2:=1/2*(x1+x/x1);
      x1:=r;
      inc(j);
     end;
     j:=3;
     if s mod 2=0 then
      begin
       l:=1;
       d[l]:=2;
       while (s mod 2=0) and (s<>2) do
        begin
         s:=s div 2;
         inc(m[2]);
        end;
      smena:=true;
      end;
     while (s>j) and (j<=round(x2)) do
      if s mod j=0 then
       begin
        if smena then begin smena:=false;inc(l);end;
        d[l]:=j;
        s:=s div j;
        inc(m[j]);
       end
      else begin inc(j,2);smena:=true;end;
     inc(l);
     d[l]:=s;
     inc(m[s]);
     s:=1;
     for j:=1 to l do
      begin
       if m[d[j]]<>0 then s:=s+m[d[j]];
       m[d[j]]:=0;
       d[j]:=0;
      end;
     k[i]:=s;
    end;
  end;
  for i:=1 to t do
   writeln(k[i]);
end.

I've got AC (but only with dirty cheatings).

Edited by author 21.10.2010 03:31

1229600
17:52:56 3 Jul 2006
wangyin
1126
Pascal
Accepted
0.5
330 KB
Funny......

maybe some special character which is not "." or "*"?

These tests have helped me to solve the problem.

Test #1
input:
A
1
1 1
output:
0

Test #2
input:
Aa
3
1 1
1 2
2 2
output:
010

Test #3
input:
AacCAa
5
1 6
1 2
1 4
2 3
5 6
output:
01001

Test #4
input:
AaCAac
6
1 6
1 2
3 6
4 5
2 3
1 5
output:
111100

Please, give me some tests to solve this problem ;)

Our mistake was following: we used wrong function of matrix multiplication. Our function had three parameters FUNC(a, b, c). We call FUNC(x, y, x) to calculate x = x * y; but there is mistake, because when we calculate resulting matrix we should use constant "x" and "y", but first source parameter "x" is the same as third resulting parameter "x". "X" is changing during calculation and so we change also source parameter.
Right decision is to use additional variable to store "x"

t = x;
FUNC(t, y, x);

That's all

example:
\""\para\""

what's right output?

I need help with code 1409:

#include <stdio.h>
int main()
{
    int a,b,c,d;
    scanf("%d%d",&a,&b);
    printf("%d\n",c=10-a,d=10-b, c+d<10);
    scanf("%d%d",&c,&d);
    getchar(), getchar();
    return 0;
}

I need help with code 1409:

#include <stdio.h>
int main()
{
    int a,b,c,d;
    scanf("%d%d",&a,&b);
    printf("%d\n",c=10-a,d=10-b, c+d<10);
    scanf("%d%d",&c,&d);
    getchar(), getchar();
    return 0;
}

Confusing question: HOW ARE THE BLANKS CHOSEN?

Let's see example
2 3 4 - initial blanks
I choose blanks 2 and 4 - they are different so I cut off piece of length 2 from the long blank (4).
After that we have blanks: 2 3 2 - two blanks coincide so one is eliminated: 2 3 they are different so we cut of blank with length 3. We have: 2 1. Cut again we have 1 1. Eliminate one blank and we get a result - 1.

You may try other ways to choose blank - but you will get same answer. So,
NO MATTER HOW TO CHOOSE 2 blanks.

Moral: What is not said does not matters ;)

By the way algorithm is very very easy... try to think

don't use longint. do it using int64

I know English bad and ask the question in Russian too.

1. I got AC with using "*2": ET = 0.015 s;
2. I replaced "*2" to "shl 1": ET = 0.001 s;
3. I tried to replace "(n-1)" to "Prev(n)": ET = 0.015 s;
4. I returned "(n-1)" instead "Prev(n)": ET = 0.015 s;

Code #4 and code #2 are identical but ET different. Why?

In Russian.

Теперь по-русски.
Прогу написал, послал -- ответ положительный, время выполнения 0,015 сек.
Заменил умножение на 2 смещением влево на 1: время выполнения 0,001 сек.
Заменил вычитание на Pred(n): Время опять 0,015;
Вернул все как было, а время выполнения не поменялось. Почему, кто скажет.

Edited by author 02.10.2010 21:48

Why WA15?
I use binarysearch in the range [-2*10^9;2*10^9]