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.