Let's see that for every number k which is an integer power of 2, k mod 3 = 1, or 2.
(To be exact - table of (k mod 3) for {1,2,4,8,16,32,64...
is: {1,2,1,2,1,2,1,2,1,2,...

So, you can see that it is impossible to get from (mod 3 = 0) position to another (mod 3 = 0) position in one move.
Let n (number of remaining stones) = 3. The first player can make x=1 or x=2 move, but now the second player can counter by making 3-x move and win. So n=3 is a losing position.

We have proven that it is impossible to get from one number of stones divisible by 3 to another number of stones divisible by 3 in one move. We can now prove, by the rule of mathematical induction, that every position with n divisible by 3 is a losing position.
Let's look - if the first player begins from position with 3*x stones, after any move the opponent can counter by '1' or '2' move and create another position with lesser number of stones, also divisible by 3.

In the second case, when first player begins from position with number of stones not divisible by 3 (so n%3=1 or n%3=2), _he_ can now make a '1' or '2' move which will create a (mod 3 = 0) position (which was proven to be losing) for his opponent. Thus, due to arbitrarity of n, any (n mod 3 != 0) position can create a losing position for the second player in one move, so it is a winning position for the first player.

Now I got it, in 3rd step player picks up 2 stones. Then it guarnatees win of 1st player.

for 12 the answer is 2
is it true

yes.
input : 3
output: 2

good luck!

Really...
Probably it's another one problem with easy tests.

PS:
На acm.timus.ru много таких задач,
тесты требуют доработки :(
Иногда проходят ну очень странные решения :)

все правильно выйграет второй, по 1 камню тоже можно брать

#include <stdio.h>
int main(){
char ch;
int sumall=0;
while (!feof(stdin)) {scanf("%c",&ch);
if (feof(stdin)) break;
if (ch>='0' && ch<='9')sumall+=(ch-'0');}<---check at this :)
sumall%=3;
printf("%d",((sumall==0)+1));
if (sumall) printf("\n%d",sumall);
return 0;
}

in each turn you must take x stones which
x is a member of S= {1, 2^1, 2^2, ...}

sorry for my poor english.

while(stream stdin hasn't finished)

>

>

1146 is simple DP. Just precount the sums of rectangles.

I agree with U about P1180!
10 lines and AC.
200 bytes
;)

I understood my mistake. Sorry.

yes, i think this solution i right

> Here is my source!
>
> #include <stdio.h>
> #include <io.h>
> int main(){
> char ch;
> int sumall=0;
> while (!feof(stdin)) {scanf("%c",&ch); if (feof(stdin))
> break;sumall+=ch-48;}
> //printf("s=%d",sumall);
> printf("%d\n",(sumall%3==0)+1);
> if (sumall%3) printf("%d",sumall%3);
> return 0;
> }
>
> Thanks!!!!