an infinite sequence_infinite sequence例子-优快云博客

探讨了一个数学问题，定义了一个无限序列A，其中A0=1，对于所有i>=1，Ai=A[i/p]+A[i/q]。通过递归算法结合缓存策略解决了该问题，确保了即使在大数情况下也能高效计算。

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)

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.

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