## Analysis of problem "A. What is for dinner?"

The solution of the problem is rather trivial. It was needed to make an array, where for each row of teeth the value of residual viability of the sickest thooth in this row would have kept (sickest tooth in the row is called the one with the lowest residual viability).

Thus we define for each row of teeth the maximum number of crucians, which Valery able to eat, using this row (Valeria can not eat more crucians, because the residual viability of the sickest tooth will become negative).

Knowing these values, you just need to sum them and to give the minimum of the sum and total amount of crucians in Valerie's portion for dinner as answer.