The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,
F(0) = 0, F(1) = 1
F(N) = F(N - 1) + F(N - 2), for N > 1.
Given N, calculate F(N).
Example 1:
Input: 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
Example 2:
Input: 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
Example 3:
Input: 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
Note:
0 ≤ N ≤ 30
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/fibonacci-number
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这里我们采用动态规划的方式
我们发现斐波那契数列的转移方程如下：
F(0) = 0, F(1) = 1
F(N) = F(N - 1) + F(N - 2),
所以我们可以用转移方程的方式去做这道题
unsigned long dp[31] = {0};
int fib(int N){
if(N == 0) return 0;
if(N == 1) return 1;
dp[0] = 0;
dp[1] = 1;
for(int i = 2; i <= N; i++)
dp[i] = dp[i-2] + dp[i-1]; //每一个值都是上两个值的和。
return dp[N];
}
