暴力枚举即可

先看题，要求是要2023年的日期，先扫一扫能组成2023的最小长度为4的序列，这个序列就用来构造年份。很明显，只需要枚举后四个数字组成的序列是否日期合法就行了。扫一眼下来，我们只需要枚举8,5,1,6,3,4,6,7,0,7,8,2,7,6,8,9,5,6,5,6,1,4,0,1,0,0,9,4,8,0,9,1,2,8,5,0,2,5,3,3段即可。

参考代码

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    unordered_set<i64> s;
    const int a[]{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    vector<i64> arr{8,5,1,6,3,4,6,7,0,7,8,2,7,6,8,9,5,6,5,6,1,4,0,1,0,0,9,4,8,0,9,1,2,8,5,0,2,5,3,3};
    int n = arr.size(), ans{};
    for(int i1 = 0; i1 < n - 3; ++i1)
        for(int i2 = i1 + 1; i2 < n - 2; ++i2)
            for(int i3 = i2 + 1; i3 < n - 1; ++i3)
                for(int i4 = i3 + 1; i4 < n; ++i4) {
                    int mouth = arr[i1] * 10 + arr[i2];
                    int day = arr[i3] * 10 + arr[i4];
                    if(mouth > 12 or mouth <= 0 or day > a[mouth] or day <= 0)
                        continue;
                    s.emplace(mouth * 100 + day);
                }
    cout << s.size();

    return 0;
}