ice: Assume that more than one Rx queue is rare in ice_napi_poll

ice: Assume that more than one Rx queue is rare in ice_napi_poll

Currently we divide budget by the number of Rx queues per Rx ring
container in ice_napi_poll even if there is only 1. This is an
unnecessary divide for the normal case of 1 Rx ring per Rx ring
container. Fix this by using an unlikely() call in the case where we
actually need to divide.

Also, we will always set budget_per_ring even if there are no Rx rings
in the Rx ring container so we don't need to initialize it to 0.

Signed-off-by: Brett Creeley <brett.creeley@intel.com>
Tested-by: Andrew Bowers <andrewx.bowers@intel.com>
Signed-off-by: Jeff Kirsher <jeffrey.t.kirsher@intel.com>