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552.student-attendance-record-ii

Statement

Metadata

可以用字符串表示一个学生的出勤记录，其中的每个字符用来标记当天的出勤情况（缺勤、迟到、到场）。记录中只含下面三种字符：

  • 'A'：Absent，缺勤
  • 'L'：Late，迟到
  • 'P'：Present，到场

如果学生能够 同时 满足下面两个条件，则可以获得出勤奖励：

  • 按 总出勤 计，学生缺勤（'A'）严格 少于两天。
  • 学生 不会 存在 连续 3 天或 连续 3 天以上的迟到（'L'）记录。

给你一个整数 n ，表示出勤记录的长度（次数）。请你返回记录长度为 n 时，可能获得出勤奖励的记录情况 数量 。答案可能很大，所以返回对 109 + 7 取余 的结果。

 

示例 1：

输入：n = 2
输出：8
解释：
有 8 种长度为 2 的记录将被视为可奖励：
"PP" , "AP", "PA", "LP", "PL", "AL", "LA", "LL" 
只有"AA"不会被视为可奖励，因为缺勤次数为 2 次（需要少于 2 次）。

示例 2：

输入：n = 1
输出：3

示例 3：

输入：n = 10101
输出：183236316

 

提示：

  • 1 <= n <= 105

Metadata

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 109 + 7.

 

Example 1:

Input: n = 2
Output: 8
Explanation: There are 8 records with length 2 that are eligible for an award:
"PP", "AP", "PA", "LP", "PL", "AL", "LA", "LL"
Only "AA" is not eligible because there are 2 absences (there need to be fewer than 2).

Example 2:

Input: n = 1
Output: 3

Example 3:

Input: n = 10101
Output: 183236316

 

Constraints:

  • 1 <= n <= 105

Solution

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const int mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
    return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
    return a > b ? a = b, 1 : 0;
}

template <typename T>
inline void chmod(T &a, const T &b) {
    a += b;
    if (a > mod) {
        a -= mod;
    }
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif
// head

const int N = 1e5 + 10;
int f[N][3][4];

class Solution {
public:
    int checkRecord(int n) {
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j < 3; j++) {
                for (int k = 0; k < 4; k++) {
                    f[i][j][k] = 0;
                }
            }
        }

        f[0][0][0] = 1;

        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < 2; j++) {
                for (int k = 0; k < 3; k++) {
                    chmod(f[i][j][0], f[i - 1][j][k]);

                    if (j) {
                        chmod(f[i][j][0], f[i - 1][j - 1][k]);
                    }

                    if (k) {
                        chmod(f[i][j][k], f[i - 1][j][k - 1]);
                    }
                }
            }
        }

        int res = 0;
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 3; j++) {
                chmod(res, f[n][i][j]);
            }
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif
最后更新: October 11, 2023
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