A geometric progression is a sequence of integers b1, b2, b3, …, where for each i > 1, the respective term satisfies the condition b_{i} = b_{i-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 |b_{i}| <= 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.

Examples:

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.

**Approach:**

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:

// CPP program to find number of terms // in Geometric Series #include <bits/stdc++.h> using namespace std; // A map to keep track of the bad integers map<int, bool> mapp; // Function to calculate No. of elements // in our series void progression(int b1, int q, int l, int m, int bad[]) { // Updating value of our map for (int i = 0; i < m; i++) mapp[bad[i]] = 1; // if starting value is greate // r than our given limit if (abs(b1) > l) cout << "0"; // if q or starting value is 0 else if (q == 0 || b1 == 0) { // if 0 is not a bad integer, // answer becomes inf if (mapp[0] != 1) cout << "inf"; // if q is 0 and b1 is not and b1 // is not a bad integer, answer becomes 1 else if (mapp[0] == 1 && mapp[b1] != 1) cout << "1"; else // else if 0 is bad integer and // b1 is also a bad integer, // answer becomes 0 cout << "0"; } else if (q == 1) // if q is 1 { // and b1 is not a bad integer, // answer becomes inf if (mapp[b1] != 1) cout << "inf"; else // else answer is 0 cout << "0"; } else if (q == -1) // if q is -1 { // and either b1 or -b1 is not // present answer becomes inf if (mapp[b1] != 1 || mapp[-1 * b1] != 1) cout << "inf"; else // else answer becomes 0 cout << "0"; } else // if none of the above case is true, // simpy calculate the number of // elements in our series { int co = 0; while (abs(b1) <= l) { if (mapp[b1] != 1) co++; b1 *= 1LL * q; } cout << co; } } // driver code int main() { // starting value of series, // number to be multiplied, // limit within which our series, // No. of bad integers given int b1 = 3, q = 2, l = 30, m = 4; // Bad integers int bad[4] = { 6, 14, 25, 48 }; progression(b1, q, l, m, bad); return 0; }

Output:

3

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