Topcoder SRM 413 (Div 2) 1000.InfiniteSequence

Problem Statement

 

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Let's consider an infinite sequence A defined as follows:

• A₀ = 1;
• A_i = 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:InfiniteSequenceMethod:calcParameters:long long, int, intReturns:long longMethod signature:long long calc(long long n, int p, int q)(be sure your method is public)

Limits

 Time limit (s):2.000Memory 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

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

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!