#include <cstdio>

int ncall;
int call[100];

void solve(int K)
{
    if (!K) return;

    if (K % 2 == 0) {
        call[ncall] = ++ncall;
        if (K > 2) solve(K / 2);
    } else {
        call[ncall++] = -1;
        solve(K - 1);
    }
}

int main( void )
{
//    freopen( "p1278.in", "r", stdin );

    int K;
    scanf( "%d", &K );

    if (K > 1) solve(K);
    for (int i = 0; i < ncall; i++)
        printf( "CALL %d\n", call[i] < 0 ? ncall : call[i] );
    printf( "BELL&RET\n" );

    return 0;
}

Nice problem! The only hint you need is that the program written below will always return 2^N.

0. CALL 1
1. CALL 2
...
N-1. CALL N
N. BELL&RET

1.Amount of calls "BELL&RET" must be equal to K.
2.Some tests, to check & understand problem :)

1
BELL&RET

2
CALL 1
BELL&RET

3
CALL 2
CALL 2
BELL&RET

4
CALL 1
CALL 2
BELL&RET

5
CALL 3
CALL 2
CALL 3
BELL&RET

6
CALL 1
CALL 3
CALL 3
BELL&RET

7
CALL 4
CALL 2
CALL 4
CALL 4
BELL&RET

here is my code:

[code deleted]

Edited by moderator 28.07.2006 10:35