Correct : C

Given:
Variables x, y, q, and r are of type unsigned int.
Code fragment:
while(r >= y)
{
  r = r - y;
  q = q + 1;
}
To find:
The initial condition on the variables before the loop starts to ensure termination with x == (y × q + r).
Solution:
Let me trace the values inside the loop first.
In each iteration of the loop, r decreases by y and q increases by 1.
Let q₀ and r₀ be the values before entering the loop.
After k iterations, the values become:
q = q₀ + k
r = r₀ - k × y
Now let me evaluate the expression y × q + r:
y × q + r = y × (q₀ + k) + (r₀ - k × y)
Multiplying and opening the brackets gives:
y × q + r = y × q₀ + k × y + r₀ - k × y = y × q₀ + r₀
So the quantity (y × q + r) is a loop invariant, which always stays equal to the initial value (y × q₀ + r₀).
We want the final state to satisfy:
x == y × q + r
Therefore, the initial values must satisfy:
x = y × q₀ + r₀
If we set q₀ = 0, then we immediately get:
r₀ = x
Now let me check loop termination.
If y = 0, r >= 0 will always be true since r is unsigned int, and r - 0 stays unchanged, so the loop will never terminate.
Thus, to ensure termination, we must have:
y > 0
Combining these conditions, before execution we must have:
( q == 0 ) && ( r == x ) && ( y > 0 )
Answer:
(C) ( q == 0 ) && ( r == x ) && ( y > 0 )