hihoCoder#1239 Fibonacci

#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：

 #include<iostream>
 #include<unordered_map>
 using namespace std;
 int fib[];
 const int m = ;
 void init() {
     int a = , b = , c = ;
     fib[] = ;
     fib[] = ;
     for (int i = ; i <= ; ++i) {
         c = a + b;
         fib[i] = c;
         a = b;
         b = c;
     }
 }
 int findPos(int x) {
     for (int i = ; i <= ; ++i) {
         if (fib[i] == x) {
             return i;
         }
     }
     return -;
 }

 int n;
 int nums[];
 long long dp[] = {};
 int main() {
     init();
     cin >> n;
     long long result = ;
     for (int i = ; i < n; ++i) {
         cin >> nums[i];
     }
     int first = -, second = -;
     for (int i = ; i < n; ++i) {
         if (nums[i] == ) {
             first = i;
             for (int j = i + ; j < n; ++j) {
                 if (nums[j] == ) {
                     second = j;
                     break;
                 }
             }
             break;
         }
     }
     if (first != -) {
         dp[] = ;
         result += ;
     }
     if (second != -) {
         dp[] = ;
         dp[] ++;
         result += ;
     }
     if (second == -) {
         cout << result << endl;
         return ;
     }

     for (int i = second + ; i < n; ++i) {
         if (findPos(nums[i]) == - ) {
             continue;
         }
         if (nums[i] == ) {  //1单独处理
             dp[] += dp[];
             dp[]++;
             result += dp[];
             dp[] %= m;
             result %= m;
             continue;
         }
         dp[findPos(nums[i])] += dp[findPos(nums[i]) - ];
         result += dp[findPos(nums[i]) - ];
         dp[findPos(nums[i])] %= m;
         result %= m;
     }
     cout << result << endl;
 }