# 1137.n-th-tribonacci-number

## Statement

• Difficulty: Easy
• Tag: `记忆化搜索` `数学` `动态规划`

T0 = 0, T1 = 1, T2 = 1, 且在 n >= 0 的条件下 Tn+3 = Tn + Tn+1 + Tn+2

``````输入：n = 4

T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4
``````

``````输入：n = 25

``````

• `0 <= n <= 37`
• 答案保证是一个 32 位整数，即 `answer <= 2^31 - 1`

• Difficulty: Easy
• Tag: `Memoization` `Math` `Dynamic Programming`

The Tribonacci sequence Tn is defined as follows:

T0 = 0, T1 = 1, T2 = 1, and Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0.

Given `n`, return the value of Tn.

Example 1:

``````
Input: n = 4
Output: 4
Explanation:
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4
``````

Example 2:

``````
Input: n = 25
Output: 1389537
``````

Constraints:

• `0 <= n <= 37`
• The answer is guaranteed to fit within a 32-bit integer, ie. `answer <= 2^31 - 1`.

## Solution

``````#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif

class Solution {
public:
int tribonacci(int n) {
auto f = vector<ll>(40);
f[1] = 1;
f[2] = 1;

for (int i = 3; i < 40; i++) {
f[i] = f[i - 1] + f[i - 2] + f[i - 3];
}

return f[n];
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````