hihocoder#1239 Fibonacci

Posted 2020-08-08 wangxiaobao的博客

#1239 : Fibonacci

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

Given a sequence {an}, how many non-empty sub-sequence of it is a prefix of fibonacci sequence.

A sub-sequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

The fibonacci sequence is defined as below:

F1 = 1, F2 = 1

Fn = Fn-1 + Fn-2, n>=3  （微软2016年秋招第三题）

输入

One line with an integer n.

Second line with n integers, indicating the sequence {an}.

For 30% of the data, n<=10.

For 60% of the data, n<=1000.

For 100% of the data, n<=1000000, 0<=ai<=100000.

输出

One line with an integer, indicating the answer modulo 1,000,000,007.

样例提示

The 7 sub-sequences are:

{a2}

{a3}

{a2, a3}

{a2, a3, a4}

{a2, a3, a5}

{a2, a3, a4, a6}

{a2, a3, a5, a6}

样例输入

6
2 1 1 2 2 3

样例输出

7

 

分析：

题意就是找到给定序列中斐波那契子序列的个数。

1. 首先想到的就是动态规划，dp[i]表示以i结尾的斐波那契子序列，然后每次变量j (0...i-1)更新i。

但是这样时间复杂度是O(n^2)，数据量10^6，肯定是超时的。

2. 考虑优化，每次更新只与斐波那契数列中的元素结尾的有关，没必要dp开那么大，并且从头遍历。

所以可以把dp存成vector<pair<int,int>>，一个表示值，一个表示以此结尾的fib序列个数。然后每次变量vector即可。

但是很遗憾，还是超时了。。。（当数组中斐波那契数列中的数存在很多时，依然是个O(n^2)）。

3. 继续优化，其实每个遍历到每个数在斐波那契序列中，更新结果时，只与其在斐波那契序列中前一个数结尾fib序列个数有关。

所以可以把dp[i]考虑存储为以fib[i]结尾的斐波那契子序列个数，100000以内斐波那契数只有25个，所以时间复杂度O(25n) = O(n)，就可以了。

注意： 用long long存result防止溢出；记得mod 1000000007

代码：

 1 #include<iostream>
 2 #include<unordered_map>
 3 using namespace std;
 4 int fib[26];
 5 const int m = 1000000007;
 6 void init() {
 7     int a = 1, b = 1, c = 2;
 8     fib[1] = 1;
 9     fib[2] = 1;
10     for (int i = 3; i <= 25; ++i) {
11         c = a + b;
12         fib[i] = c;
13         a = b;
14         b = c;
15     }
16 }
17 int findPos(int x) {
18     for (int i = 1; i <= 25; ++i) {
19         if (fib[i] == x) {
20             return i;
21         }
22     }
23     return -1;
24 }
25 
26 int n;
27 int nums[1000000];
28 long long dp[26] = {0};
29 int main() {
30     init();
31     cin >> n;
32     long long result = 0;
33     for (int i = 0; i < n; ++i) {
34         cin >> nums[i];
35     }
36     int first = -1, second = -1;
37     for (int i = 0; i < n; ++i) {
38         if (nums[i] == 1) {
39             first = i;
40             for (int j = i + 1; j < n; ++j) {
41                 if (nums[j] == 1) {
42                     second = j;
43                     break;
44                 }
45             }
46             break; 
47         }
48     }
49     if (first != -1) {
50         dp[1] = 1;
51         result += 1;
52     }
53     if (second != -1) {
54         dp[2] = 1;
55         dp[1] ++;
56         result += 2;
57     }
58     if (second == -1) {
59         cout << result << endl;
60         return 0;
61     }
62     
63     for (int i = second + 1; i < n; ++i) {
64         if (findPos(nums[i]) == -1 ) {
65             continue;
66         }
67         if (nums[i] == 1) {  //1单独处理 
68             dp[2] += dp[1];
69             dp[1]++;
70             result += dp[1];
71             dp[2] %= m;
72             result %= m;
73             continue;
74         }
75         dp[findPos(nums[i])] += dp[findPos(nums[i]) - 1];
76         result += dp[findPos(nums[i]) - 1];
77         dp[findPos(nums[i])] %= m;
78         result %= m;
79     }
80     cout << result << endl;
81 }