1137.n-th-tribonacci-number

Statement

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Metadata

• Link: 第 N 个泰波那契数
• Difficulty: Easy
• Tag: 记忆化搜索 数学 动态规划

泰波那契序列 T_n 定义如下： 

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

给你整数 n，请返回第 n 个泰波那契数 T_n 的值。

 

示例 1：

输入：n = 4
输出：4
解释：
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4

示例 2：

输入：n = 25
输出：1389537

 

提示：

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

Metadata

• Link: N-th Tribonacci Number
• Difficulty: Easy
• Tag: Memoization Math Dynamic Programming

The Tribonacci sequence T_n is defined as follows: 

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

Given n, return the value of T_n.

 

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

C++

#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
// head

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

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最后更新: October 11, 2023