#include stdioh#define SQRX XXvoid main int a = 16k = 2 m = 1; a = SQRk+m SQRk+m; printfdna;

The output of this code is 16.

Explanation: The macro SQR(X) is defined as X*X, which means that any occurrence of SQR(k+m) in the code will be replaced by (k+m)*(k+m).

So, the line a /= SQR(k+m) / SQR(k+m); will be evaluated as follows:

1. First, k+m evaluates to 3 (since k=2 and m=1).
2. SQR(k+m) is replaced by (k+m)*(k+m), which gives (3)*(3) = 9.
3. So, the expression becomes a /= 9/9.
4. 9/9 evaluates to 1.
5. Finally, a /= 1 is equivalent to a = a/1, which means that a remains unchanged and its value is still 16.

Therefore, the output of the code is 16.