First, let's remember for each dish how many other \bf{unplanted} people like it, let for dish $$$i$$$ this value will be equal to $$$b_i$$$.

The following greedy solution works: let's try to take a dish $$$j$$$ such that $$$b_j$$$ is the maximum of the array $$$b$$$, consider another free table with a maximum capacity of $$$k$$$. Then you can put $$$\min(b_j, c_k)$$$ guests at table $$$k$$$, and change $$$b_j := b_j - \min(b_j, c_k)$$$. After that, you need to continue the algorithm, if there are still free tables and unsettled guests. If there are unseated guests at the end, we will seat them at any tables — they will not be satisfied, that is, we need to minimize the number of free seats.

Let's prove this solution:

1) Obviously, it is not profitable to put dish $$$i$$$ on the table and at the same time not seat the maximum possible number of people who have dish $$$i$$$ as their favorite.

2) It is profitable to take the maximum $$$b_i$$$ and seat $$$k$$$ at the table with the maximum $$$c_k$$$. Let's assume that we seated at table $$$k$$$ not $$$b_i$$$, but another $$$b_j \le b_i$$$. Then if:

• $$$b_i, b_j \le c_k$$$; then it is obvious that it is better to seat $$$b_i$$$, because there will be $$$c_k - b_i \le c_k - b_j$$$ free seats.

• $$$b_j \le c_k \le b_i$$$; then there will be $$$c_k - b_j$$$ empty seats at table $$$k$$$, but if $$$b_i$$$ were seated, there would be $$$0$$$, so this option is not better.

• $$$c_k \le b_i, b_j$$$; then it would be no worse to plant $$$b_i$$$, since $$$b_j - c_k \le b_i - c_k$$$. That is, at the next iterations of the algorithm, it is better for us to have a maximum as large as possible; when choosing $$$b_i$$$, the maximum will be no less than when choosing $$$b_j$$$.

That is, taking the maximum $$$b_i$$$ is no worse than taking another $$$b_j \le b_i$$$, so we are sure that at each iteration of the algorithm we seat the maximum possible number of guests.

Also I haven't been in SIS winter