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Leecode＜每日一题＞学生出勤记录 II题目链接思路：动态规划，dp[i][0]，表示n == i，缺勤次数为0时的可能情况数。dp[i][1]，表示n == i，缺勤次数为1时的可能情况数。class Solution {public: int checkRecord(int n) { long long dp[100010][2]{}; dp[1][0] = 2,dp[1][1] = 1; dp[2][0] = 4,dp[2][1]

Leecode＜每日一题＞出界的路径数题目链接思路：动态规划，dp[i][j][step] 表示在位置[i,j]处，可移动距离为step时，出界的路径数。class Solution {public:int dp[51][51][51]{};vector<vector<int>> dir= { {1,0},{-1,0},{0,1},{0,-1} }; int findPaths(int m, int n, int maxMove, int startRow, i

Leecode＜每日一题＞最长回文子序列题目链接思路：动态规划dp[i][j]表示下标i到j的区间内，最大的回文子序列的长度。动态转移方程：dp[i][end] = dp[i + 1][end - 1] + 2; //区间首位相同的转移方程dp[i][end] = max(dp[i][end - 1],dp[i+1][end]); //区间首位不同的转移方程class Solution {public: int longestPalindromeSubseq(string s

Leecode＜每日一题＞等差数列划分 II - 子序列题目链接思路：动态规划+哈希表class Solution {public:typedef long long ll; int numberOfArithmeticSlices(vector<int>& nums) { auto tab = vector<unordered_map<int, int>>(nums.size() + 1); int sum = 0; for (int

#include"stdio.h"#include"algorithm"#include"iostream"#include"math.h"#include"string.h"using namespace std;int f[51][40000];int main(){ memset(f, 0, sizeof(f)); int N; int ai[51],b

题目网址：http://acm.fzu.edu.cn/problem.php?pid=2234#include"stdio.h"_int64 dp[105][105][200];_int64 max(_int64 a,_int64 b){ return a>=b?a:b;}int main(){ _int64 const min=1e18; int i,j,k,n

Let’s play a stone removing game. Initially, n stones are arranged on a circle and numbered 1,…,n clockwise (Figure 1). You are also given two numbers k and m. From this state, remove stones one by one