All arrows of pills are 3 cells long, so their tens digit can only be 1 or 2.

By checking the parity of row 9, we can see that:

• The sum of r9c2~r9c5 must be even

• The sum of r9c6~r9c9 must be even

• Therefore, r9c1 must be odd, which can only be 1.

And r4c1 now must be 2.

The arrow on r4c1 can only be 23 (3 is the minimum digit left in the column) or 24 (24 = 7 + 8 + 9, which are the maximal digits we can use).

In this case, the other 3 cells left must sum to 14 or 16.

We can also get r1c2, r3c2, r1c7.

For the same reason, The arrow on r3c2, r1c7 must be 23 or 24, and the digits on the arrows must be 689 or 789.

The arrows in r4c1, r3c1 both contain 8 and 9, so we have used two 8s and two 9s from column 1 and 2. The 8 and 9 in box 1 can only be on column 3.

• Therefore, there is a 89 pair in r2c3, r3c3.

r{4,5,6}c3 can only be 1567.

The arrow on r7c3 can only be 6 or 7.

Now we found a 1567 quadruple in column 3.

Now that we have a 1567 quadruple in column 3, r1c3 must be 4.

We can put in a lot of digits in c1, c2, c3.

The arrow on r2c6 is very restricted now.

• To get the maximum value, r2c3 can be 9, r2c{4,5} can only be 45, the sum is 18.

• Therefore, r2c6 can only be 1.

r8c6 now must be 2, r9c6 can only be 3 or 4, with 689 or 789 on the arrow.

The r9c1 arrow can only be 13 or 14. And we can get a few more digits.

r8c9 is a naked single, it can only be 4.

The arrow on r4c9 is very restricted now.

• The minimum of r2c9, r3c9 is 56 pair, and 567 adds up to 18.

• We can't increase the sum by just 1, therefore, 19 is impossible.

• The arrow must be 18 = 5 + 6 + 7.

In box 2, 2 can only be in r2c4 or r2c5.

Therefore, r2c3 must be 9. (Otherwise we will have 8+2+x = 10+x, and we will need two x in row 2)

r2c7 can only be 4, 8 or 9. But the maximum value we can put on the arrow is 9+2+5=16.

Therefore, the only valid value is 14. And r2c{4,5} is a 23 pair.

We can get more digits.

• There is a 235 triple in r7c{7,8,9}.

There is a 69 pair in column 5.

In box 5, 9 must be in r6c6.

We can resolve the 689 in box 2, box 7 and box 8.

r2c5, r6c9 cannot be both 2, otherwise we will rule out 2 from both r6c4, r5c5.

Therefore, at least one of r2c5, r6c9 is 3. This rules out 3 from the candidate of r6c5.

Therefore, r5c4 must be 3.

The rest of the solve should be very straightforward.