【强训笔记】day4

NO.1

思路：利用滚动数组，迭代一个Fibonacci数列，给出三个值进行循环迭代，当n<c时，说明n在b和c之间，这里只需要返回c-n和n-b的最小值就可以了。

代码实现：

#include<iostream>

using namespace std;

int 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;
}

NO.2

代码实现：

class Solution
{
	int m, n;
	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 = j + dy[k];
			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;
	}
};

NO.3

思路：dp线性数组，dp11=1,在根据状态方程dpij = dpi - 1j + dpi - 1j - 1进行求解。

#include<iostream>

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

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