The first thing we can notice is that if a printer has $u$ units of ink left of a given color, we cannot use more than $u$ units of that color. Moreover, this is the only restriction imposed by that value. So, we can summarize the input by saying we cannot use more than $C=min(C_1,C_2,C_3)$ units of cyan ink, $M=min(M_1,M_2,M_3)$ units of magenta ink, $Y=min(Y_1,Y_2,Y_3)$ units of yellow ink, or $K=min(K_1,K_2,K_3)$ units of black ink.

If $C+M+Y+K<{10}^6$ then the case is impossible and we are done. Otherwise, we may need to use lower amounts of each color. We can simply go one color at a time, lowering the amounts of ink until we make the sum exactly ${10}^6$. Doing it one unit at a time works, but it is very slow. We can do better: in the same way as before, we can consider all the colors one at a time. Let $S$ be the sum of the current amount of ink for the $3$ colors not currently under consideration. If $S\ge {10}^6$, we can simply set the amount of the current color to $0$ and continue with the next one. If $S<{10}^6$ we can lower the current color to ${10}^6-S$ and finish immediately. This works because at all times we maintain the invariant that the total amount of ink we are considering is at least ${10}^6$ units.