In python 3.6 extra print() in the end of the program gives you WA2. Therefore you should output
"8
 1 3 4 2 0 1 3 2"
instead of
"8
 1 3 4 2 0 1 3 2
 "

Subj

Edited by author 24.08.2014 16:33

Please, what is the input in test 1? I'm almost sure that I solved the problem right, but I still have WA1. It seems like the format of output doesn't work..

My solution use emulation with some heuristic conditions (for avoiding deadlock). Solution is o(n).

Some test:
1
output
3
0 2 1
--------
2
output
8
1 3 4 2 0 1 3 2
--------
3
output
15
2 4 5 3 1 0 2 4 6 5 3 1 2 4 3
--------
4
output
24
3 5 6 4 2 1 3 5 7 8 6 4 2 0 1 3 5 7 6 4 2 3 5 4
--------
5
output
35
4 6 7 5 3 2 4 6 8 9 7 5 3 1 0 2 4 6 8 10 9 7 5 3 1 2 4 6 8 7 5 3 4 6 5

    Please tell me.
    Thanks.

i am on the 5th place of this problem ranking!!! (it's my record :)). if the sorting was also on the memory then i would be on 2nd place - and all of it i did at 5 o'clock, at night, on Pascal.
 0.031 - 154kb

Is there more simple, mathematical way to solve this problem other than actually play the game with some smart heuristics (what I did in my solution)?

#include <stdio.h>

int main(void)
{
  int n,m,i,p,d;
  char f[999];
  scanf("%d%d",&n,&m);
  for(p=0;p<n;p++){
    f[p]='W';
    f[n+1+p]='B';
  }
  f[n]='_';
  for(i=0;i<m;i++){
    scanf("%d",&p);
    if(p<0||2*n<p||f[p]=='_'){
      printf("error: strange number (%d)\n",p);
      return 1;
    }
    d=(f[p]=='W'?1:-1);
    if(p+d<0||2*n<p+d){
      printf("error: junp to out (%d->%d)\n",p,p+d);
      return 1;
    }
    if(f[p+d]!='_')d*=2;
    if(p+d<0||2*n<p+d||f[p+d]!='_'){
      printf("error: jump to not empty (%d->%d)\n",p,p+d);
      return 1;
    }
    f[p+d]=f[p];
    f[p]='_';
    // printf("%d->%d\n",p,p+d);
  }
  for(p=0;p<=2*n;p++)
    putchar(f[p]);
  putchar('\n');
  return 0;
}


Input

4
24
3 5 6 4 2 1 3 5 7 8 6 4 2 0 1 3 5 7 6 4 2 3 5 4

Output

BBBB_WWWW

Edited by author 22.08.2007 11:51

I've submitted the following solution, and it got AC:

#include <cstdio>

int main(void)
{
    int N;
    scanf("%d", &N);
    if (N==1)
    {
        printf("3\n2 0 1\n");
    } else
    {
        printf("0\n");
    }
    return 0;
}


Probably, this is a bug in checker

What is the answear for N=2 && N=3?