The dog has been running at the speed of z, so just calculate how long the dog has been running.

The dog and the brother set out at the same time until the brother caught up with the younger brother, so it was actually to calculate the time when the brother caught up with the younger brother.

When an elder brother catches up with his younger brother, it is actually catching up with the problem.

The distance difference between two people is

X*M

, the speed difference is

Y-X

So it takes time for my brother to catch up with him.

X*M*(Y-X)

So the dog ran away.

Z*X*M*(Y-X)

If it is a C language program problem, there should be no problem.

X & gtY

This irrelevant situation. . .