class Solution {
public:/*动态规划五部曲：1、确定dp数组以及下标的含义：dp[i][j]，到dp[i][j]位置上的不同路径数量；2、确定递推公式:每次可由上边或者左边的位置移动而来，所以dp[i][j] = dp[i-1][j] + dp[i][j-1]；3、dp数组如何初始化：dp[0][i]=1，dp[i][0]=1；4、确定遍历顺序：根据递推公式，从前往后遍历，从上往下遍历；5、举例推导dp数组；*/int uniquePaths(int m, int n) {vector<vector<int>> dp(m, vector<int>(n, 0));for(int i = 0; i < m; i++) dp[i][0] = 1;for(int i = 0; i < n; i++) dp[0][i] = 1;for(int i = 1; i < m; i++){for(int j = 1; j < n; j++){dp[i][j] = dp[i-1][j] + dp[i][j-1];}}return dp[m-1][n-1];}
};

class Solution {
public:int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {int m = obstacleGrid.size();int n = obstacleGrid[0].size();vector<vector<int>> dp(m, vector<int>(n, 0));for(int i = 0; i < m; i++){if(obstacleGrid[i][0] == 1) break;dp[i][0] = 1;}for(int i = 0; i < n; i++){if(obstacleGrid[0][i] == 1) break;dp[0][i] = 1;}for(int i = 1; i < m; i++){for(int j = 1; j < n; j++){if(obstacleGrid[i][j] == 1) dp[i][j] = 0;else dp[i][j] = dp[i-1][j] + dp[i][j-1];}}return dp[m-1][n-1];}
};

class Solution {
public:int integerBreak(int n) {vector<int> dp(n+1);      //dp[i]为拆分正整数i的最大乘积；dp[2] = 1;              //dp[0]、dp[1]没有实际意义，从索引2开始初始化dp[]；for(int i = 3; i <= n; i++){for(int j = 1; j <= i/2; j++){dp[i] = max(dp[i], max((i-j)*j, dp[i-j]*j));}}return dp[n];}
};