dp[i][j]=(dp[i-1][j+1] + dp[i][j-1])%MOD;
时间: 2023-10-23 22:39:23 浏览: 28
As an AI language model, I am programmed to understand the concept of null as a value that represents the absence or lack of a value or object. In programming, null is often used to indicate that a variable or object has not been initialized or has no value. In databases, null is used to indicate the absence of a value in a field.
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dp[i]+=dp[i-j];
This statement is incomplete and lacks context, but it appears to be part of a dynamic programming algorithm. Generally, in dynamic programming, dp[i] represents the value of the optimal solution to a subproblem at index i, and dp[i-j] represents the value of the optimal solution to a subproblem at index i-j. Without more information about the specific problem and algorithm, it is difficult to provide a more detailed explanation.
以下哪个选项是编辑距离问题的动态规划转移方程 A dp[i][j]=dp[i-1][j-1] 当a[i-1]==b[j-1] dp[i][j]=min(dp[i-1][j-1]+1, dp[i][j-1]+1, dp[i-1][j]+1) 当a[i-1]!=b[j-1] B dp[i][j]=dp[i-1][j-1] 当a[i-1]!=b[j-1] dp[i][j]=min(dp[i-1][j-1]+1, dp[i][j-1]+1, dp[i-1][j]+1) 当a[i-1]==b[j-1] C dp[i][j]=dp[i-1][j-1] 当a[i-1]==b[j-1] dp[i][j]=min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) 当a[i-1]!=b[j-1] D dp[i][j]=dp[i-1][j-1] 当a[i-1]==b[j-1] dp[i][j]=min(dp[i][j-1]+1, dp[i-1][j]+1) 当a[i-1]!=b[j-1]
动态规划转移方程通常是dp[i][j]与dp[i-x][j-y]的关系,其中x和y是根据具体问题而定的。在编辑距离问题中,dp[i][j]表示将字符串a[0:i-1]转化为字符串b[0:j-1]所需的最小编辑距离。因此,答案是B。
当a[i-1]!=b[j-1]时,dp[i][j]的值可以由dp[i-1][j-1]、dp[i][j-1]和dp[i-1][j]这三个状态转移而来,分别对应于将a[i-1]替换为b[j-1]、在a[i-1]后面添加b[j-1]和在b[j-1]后面添加a[i-1]这三种编辑操作。因此,dp[i][j]=min(dp[i-1][j-1]+1, dp[i][j-1]+1, dp[i-1][j]+1)。
当a[i-1]==b[j-1]时,dp[i][j]的值可以由dp[i-1][j-1]转移而来,表示a[i-1]和b[j-1]相同,不需要编辑操作。因此,dp[i][j]=dp[i-1][j-1]。