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。
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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);
}
};