Likou 115. Different sub-sequence dp

https://leetcode-cn.com/problems/distinct-subsequences/

Idea: $d{p}_{ij}$Means $s[i:In$ the subsequence of]$t[j:]$ Number of occurrences, then initially$dp[i][m]=1$，当$s[i]=When\; t\left[j\right]$ ,$t\left[j\right]$ can choose whether to match$s\left[i\right]$ matches, so we get$dp[i][j]=dp[i+1][j+1]+dp[i+1][j]$；否则$dp\left[i\right]\left[j\right]=dp[i+1]\left[j\right]$ . To open longlong.

class Solution {
    
    
public:
    int numDistinct(string s, string t) {
    
    
        int n=s.size(),m=t.size();
        if(n<m)
            return 0;
        using ll=long long;
        vector<vector<ll>> dp(n+1,vector<ll>(m+1));
        for(int i=0;i<=n;i++)
            dp[i][m]=1;
        for(int i=n-1;i>=0;i--)
        {
    
    
            int minj=max(0,m-(n-i));
            for(int j=m-1;j>=minj;j--)
            {
    
    
                dp[i][j]=dp[i+1][j];
                if(s[i]==t[j])
                    dp[i][j]+=dp[i+1][j+1];
            }
        }
        return dp[0][0];
    }
};