Why Efficiency is always useful power out divided by power in. For an RF amplifier the useful output is the RF signal power delivered to the load, and the power in is the DC power drawn from the supply (volts times amps at the collector or plate). So efficiency = RF output power / DC input power, usually multiplied by 100 to express it as a percent; the rest of the DC power becomes heat. Typical figures are roughly 25-35 percent for Class A, 50-60 percent for Class AB, and up to 80 percent or more for Class C.
Watch out Dividing DC input by DC output is meaningless here because the amplifier's output is RF, not DC, and that ratio would be upside down anyway. Any answer that adds powers or uses reciprocals is not a ratio of output to input at all.
Efficiency = what you get out over what you pay for: RF out / DC in. Output on top, always.