Solution Notes: This problem can be solved by dynamic programming. We treat inputs with K=1 as a special case, since this involves just printing a one followed by N-1 zeros. For K at least 2, a quick back-of-the-envelope calculation shows us that the total number of digits in the answer will be at most 5000. For the two-dimensional array A[0..10][0..5000], we let A[i][j] denote the number of j-digit binary numbers (including those that start with leading zeros) with exactly i 1-bits. We can fill in this table by setting A[i][j] = A[i-1][j-1] + A[i][j-1], since a j-digit number with i 1-bits can be obtained either by appending a 0 bit to a (j-1)-digit number with i 1-bits, or by appending a 1 bit to a (j-1)-digit number with (i-1) 1-bits. Once we have filled in the table, the appropriate "traceback path" from A[K][5000] gives us the binary number we seek (taking care not to print leading zeros).

#include <stdio.h>
#define M 5000

int A[11][M+1];
int leading_zeros = 1;

void print_sol(int n,int k,int m)
{
  if (k==0 && m==1) return;
  if (k==0 || A[k][m-1] >= n) {
    if (!leading_zeros) printf ("0");
    print_sol(n,k,m-1);
  } else {
    leading_zeros = 0;
    printf ("1");
    print_sol(n-A[k][m-1],k-1,m-1);
  }
}

int main(void)
{
  int i,j,N,K;

  freopen ("cowids.in", "r", stdin);
  freopen ("cowids.out", "w", stdout);

  scanf ("%d %d", &N, &K);

  if (K==1) {
    printf ("1");
    for (i=0; i<N-1; i++) printf ("0");
    printf ("\n");
    return 0;
  }

  A[0][1] = 1;
  for (j=1; j<=M; j++) {
    for (i=0; i<=10; i++) {
      if (i==0) A[i][j] = 1;
      else A[i][j] = A[i-1][j-1] + A[i][j-1];
      if (A[i][j] > 10000000) A[i][j] = 10000000; /* avoid overflow */
    }
  }

  print_sol(N,K,M);
  printf ("\n");

  return 0;
}