Topcoder SRM 413 (Div 2) 1000.InfiniteSequence

Problem Statement
	NOTE: This problem statement contains subscripts that may not display properly if viewed outside of the applet. Let's consider an infinite sequence A defined as follows: A0 = 1; Ai = A[i/p] + A[i/q] for all i >= 1, where [x] denotes the floor function of x. (see Notes) You will be given n, p and q. Return the n-th element of A (index is 0-based).
Definition
	Class: InfiniteSequence Method: calc Parameters: long long, int, int Returns: long long Method signature: long long calc(long long n, int p, int q) (be sure your method is public)
Limits
	Time limit (s): 2.000 Memory limit (MB): 64
Notes
-	[x] denotes the floor function of x which returns the highest integer less than or equal to x. For example, [3.4] = 3, [0.6] = 0.
Constraints
-	n will be between 0 and 10^12, inclusive.
-	p and q will both be between 2 and 10^9, inclusive.
Examples
0)
	0 2 3 Returns: 1 A[0] = 1.
1)
	7 2 3 Returns: 7 A[0] = 1; A[1] = A[0] + A[0] = 2; A[2] = A[1] + A[0] = 2 + 1 = 3; A[3] = A[2] + A[1] = 3 + 2 = 5; A[7] = A[3] + A[2] = 5 + 3 = 8.
2)
	10000000 3 3 Returns: 32768
3)
	256 2 4 Returns: 89
4)
	1 1000000 1000000 Returns: 2

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My Solution

Just memory search

// BEGIN CUT HERE

// END CUT HERE
#line 5 "InfiniteSequence.cpp"
#include <string>
#include <vector>
#include <map>
//#include <cmath>
using namespace std;
class InfiniteSequence {
	public:
    map<long long, long long> d;
	long long calc(long long n, int p, int q) {
		if(n == 0) return d[0] = 1;
		d[n] = (d[n/p] != 0) ? d[n/p] : calc(n/p, p, q);
		return d[n] += (d[n/q] != 0) ? d[n/q] : calc(n/q, p, q);
	}
};

Thank you!