An attendance record for a student can be represented as a string where each character signifies whether the student was absent, late, or present on that day. The record only contains the following three characters:

• 'A': Absent.
• 'L': Late.
• 'P': Present.

Any student is eligible for an attendance award if they meet both of the following criteria:

• The student was absent ('A') for strictly fewer than 2 days total.
• The student was never late ('L') for 3 or more consecutive days.

Given an integer n, return the number of possible attendance records of length n that make a student eligible for an attendance award. The answer may be very large, so return it modulo \(10^9 + 7\).

Solution

考虑记忆化搜索，用 \(dfs(i,n,absent, late)\) 表示。

class Solution {
private:
    int mod=1e9+7;
    int dp[100002][5][5];
    
    int dfs(int i, int n, int absent, int late){
        if(absent>1 || late>=3) return 0;
        if(i==n) return 1;
        if(dp[i][absent][late]!=-1)
            return dp[i][absent][late];
        int ans=0;
        
        // absent
        ans = (ans%mod + dfs(i+1, n, absent+1, 0)%mod)%mod;
        // late
        ans = (ans%mod + dfs(i+1, n, absent, late+1)%mod)%mod;
        // pre
        ans = (ans%mod + dfs(i+1, n, absent, 0)%mod)%mod;
        
        return dp[i][absent][late]=ans;
        
    }
    
    
public:
    int checkRecord(int n) {
        memset(dp, -1, sizeof(dp));
        return dfs(0, n, 0,0);
    }
};