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A math problem related to virus
The division times of virus A: 10/0.5=20 times.

The division times of virus A: 2 * 2 * 2 *(20 times) = 1048576.

Number of times virus B increased: 10/0.5=20 times.

The number of virus B increased: 2 * (20+1) = 420,000.

There are 65438 A viruses+0048576 > 420000 B viruses.

Remaining quantity of virus A and virus B after cancellation: 420000- 1048576=(-628576).

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