The task itself is a very easy task. A very simple implementation just check if the difference between the first two equals to the difference between the second two elements. There is a little part to take care of. when one element equals to 0 we might end up dividing on 0 when we are checking if the progression is geometrical. Use this pseudocode
if(a2-a1==a3-a2)
print "arithemitc"
else
print "gemoetric"
Don't check for geometric progression
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