###### tags: `LeetCode` `Easy` `Dynamic Programming`
# LeetCode #1137 N-th Tribonacci Number
### (Easy)

泰波那契序列 Tn 定義如下： 

T0 = 0, T1 = 1, T2 = 1, 且在 n >= 0 的條件下 Tn+3 = Tn + Tn+1 + Tn+2

給你整數 n，請返回第 n 個泰波那契數 Tn 的值

---

class Solution {
public:
    int tribonacci(int n) {
        if(n==0)return 0;
        else if(n==1||n==2)return 1;
        int dp[n+1];
        dp[0]=0, dp[1]=1,dp[2]=1;
        for(int i=3;i<=n;i++){
            dp[i]=dp[i-3]+dp[i-2]+dp[i-1];
        }
        return dp[n];
    }
};