<Sicily>Fibonacci

一、题目描述

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn-1 + Fn-2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

Given an integer n, your goal is to compute the last Fn mod (10^9 + 7).

二、输入

The input test file will contain a single line containing n (n ≤ 100).

There are multiple test cases!

三、输出

For each test case, print the Fn mod (10^9 + 7).

例如：
输入：9
输出：34

四、解题思路

使用动态规划思想，大问题由小问题组成。
第3项可以由1，2项求得，
第4项可以由2，3项求得，
…
第n项可以由n - 2, n - 1求得。

五、代码

#include <iostream>using namespace std;const long long int MODE = 1000000000+7;    //当数太大是取模（题目要求）int main(){  int n;  while(cin >> n){      if(n == 0) cout << 0 << endl;      if(n == 1) cout << 1 << endl;      if(n >= 2){          int fn0, fn1, fn2;          fn0 = 0;          fn1 = fn0 + 1;          for(int i = 0; i < n - 1; i++)          {              fn2 = fn1 + fn0;              fn0 = fn1;              fn1 = fn2;          }          fn2 %= MODE;          cout << fn2 << endl;      }  }  return 0;}