A geometric progression is a sequence of integers b1, b2, b3, …, where for each i > 1, the respective term satisfies the condition bi = bi-1 * q, where q is called the common ratio of the progression.
Given geometric progression b defined by two integers b1 and q, and m “bad” integers a1, a2, .., am, and an integer l, write all progression terms one by one (including repetitive) while condition |bi| <= l is satisfied (|x| means absolute value of x).
Calculate how many numbers will be there in our sequence, or print “inf” in case of infinitely many integers.
Note: If a term equals one of the “bad” integers, skip it and move forward to the next term.
Input : b1 = 3, q = 2, l = 30, m = 4 6 14 25 48 Output : 3 The progression will be 3 12 24. 6 will also be there but because it is a bad integer we won't include it Input : b1 = 123, q = 1, l = 2143435 m = 4 123 11 -5453 141245 Output : 0 As value of q is 1, progression will always be 123 and would become infinity but because it is a bad integer we won't include it and hence our value will become 0 Input : b1 = 123, q = 1, l = 2143435 m = 4 5234 11 -5453 141245 Output : inf In this case, value will be infinity because series will always be 123 as q is 1 and 123 is not a bad integer.
We can divide our solution in different cases:
Case 1: If starting value of series is greater than the given limit, output is 0.
Case 2: If starting value of series or q is 0, there are three more cases:
Case 2.a: If 0 is not given as a bad integer, answer will become inf.
Case 2.b: If b1 != 0 but q is 0 and b1 is also not a bad integer, answer will become 1.
Case 2.c: If 0 is given as a bad integer and b1 = 0, answer will become 0.
Case 3: If q = 1 we will check if b1 is given as a bad integer or not. If it is then answer will be 0 else answer will be inf.
Case 4: If q = -1, check if b1 and -b1 is present or not, if they are present our answer will be 0 else our answer will be inf.
Case 5: If none of the above cases hold, simply run a loop for b1 till l and calculate the number of elements.
Below is the implementation of above approach:
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Improved By : mohit kumar 29