【day04】Fibonacci数列 | 单词搜索

【day04】Fibonacci数列 | 单词搜索 | 杨辉三角

北海有初拥2025-05-10 17:29

1.Fibonacci数列

解题思路：

求斐波那契数列的过程中，判断⼀下：何时n会在两个fib数之间。

复制代码

#include <bits/stdc++.h>

using namespace std;

#define int long long

signed main()
{
    int n;
    cin >> n;
    int a = 0 , b = 1 , c = 1;
    while(n > c)
    {
        a = b;
        b = c;
        c = a + b;
    }
    cout << min(c - n , n - b) << endl;
    return 0;
}

2.单词搜索

题目链接：单词搜索_牛客题霸_牛客网

解题思路：简单的深度搜索题（dfs）

复制代码

class Solution
{
    int n, m;
    bool vis[101][101] = {0};
    int dx[4] = {0, 0, 1, -1};
    int dy[4] = {1, -1, 0, 0};

public:
    bool exist(vector<string> &board, string word)
    {
        m = board.size(), n = board[0].size();
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (board[i][j] == word[0])
                {
                    if (dfs(board, i, j, word, 0))
                        return true;
                }
            }
        }
        return false;
    }

    bool dfs(vector<string> &board, int i, int j, string &word, int pos)
    {
        if (pos == word.size() - 1)
            return true;

        vis[i][j] = true;
        for (int k = 0; k < 4; k++)
        {
            int a = i + dx[k], b = dy[k] + j;
            if (a >= 0 && a < m && b >= 0 && b < n && !vis[a][b] && board[a][b] == word[pos + 1])
            {
                if (dfs(board, a, b, word, pos + 1))
                    return true;
            }
        }
        vis[i][j] = false;
        return false;
    }
};

3.杨辉三角

题目链接：

杨辉三角_牛客题霸_牛客网

解题思路：最基础的dp模型，按照规律模拟出来杨辉三⻆即可。

复制代码

#include <bits/stdc++.h>

using namespace std;

#define int long long

int dp[31][31];
signed main()
{
    int n;
    cin >> n;
    dp[1][1] = 1;
    for(int i = 2;i <= n;i ++)
    {
        for(int j = 1;j <= n;j ++)
        {
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
        }
    }

    for(int i = 1;i <= n;i ++)
    {
        for(int j = 1;j <= n;j ++)
        {
            printf("%5d", dp[i][j]);
        }
        printf("\n");
    }
    return 0;
}