an infinite sequence

Problem Statement

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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)

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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解法：

#include <fstream>#include <sstream>#include <iostream>#include <algorithm>#include <vector>#include <list>#include <set>#include <string>#include <map>using namespace std;int f(int n, int p, int q, map<int, int> & r){ static map<int, int>::iterator it; static int tmp; if(p < q) //swap p & q if p < q { tmp = p; p = q; q = tmp; } it = r.find(n); if(it == r.end()) { int i = f(n/p, p, q, r); r.insert(pair<int, int>(n/p, i)); int j = f(n/q, p, q, r); r.insert(pair<int, int>(n/q, j)); return i + j; } else return it->second;}int main(int argc, char * argv[]){ map<int, int> r; r.insert(pair<int, int>(0, 1)); r.insert(pair<int, int>(1, 2)); cout << f(1000000000000, 99999999, 256427, r); return 0;}

 

 

1. 直接递归可能导致栈溢出, 因此使用map保存中间已经计算出的结果

2. 交换p & q， 可以减少递归的深度