**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;
}
```