How can I make the sum of A's last count =26?

A and B can only quote 1, 2, 3.

That only needs to be counted before the last round, and the sum is greater than 23 and less than 25.

Only the sum before the number B =26-4=22, that is, a penultimate makes the sum =22.

Then, no matter how B reports, A can report at last, making the sum =26.

Similarly, the sum before the second reciprocal of B =22-4= 18, that is, the third reciprocal of A makes the sum = 18.

Similarly, the fourth reciprocal of A makes sum = 18-4= 14.

The fifth reciprocal of a makes the sum = 14-4= 10.

The sixth reciprocal of a makes the sum = 10-4=6.

The seventh reciprocal of A makes the sum =6-4=2.

So the seventh penultimate time of A should be the first time that A starts to count off, and the count off is 2. Assuming that the future (B counts off, A counts off) is a round, then each round only needs to add up to 4.

Conversely, from the first count, you can get the answer.

A number 2

B report 1 A report 3 and =2+4=6.

The number B is 3 A 1 and =2+4+4= 10.

The number 2 of B and the number 2 of A = 2+4+4 =14.

B number 3 A number 1 and = 2+4+4 =18.

B report 1 A report 3 and = 2+4+4+4 = 22.

The number of B is 2, and the number of A is 2 = 2+4+4+4+4 = 26.