枚举合法日期：

---

#include <bits/stdc++.h>
using namespace std;
 
int month[13] = {-1, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int date, ans1, ans2;
 
// 判断日期是否为回文
bool check1(int date) {
    string s = to_string(date);
    if (s[0] == s[7] && s[1] == s[6] && s[2] == s[5] && s[3] == s[4])
        return true;
    return false;
}
 
// 判断日期是否满足 ABABBABA 型回文
bool check2(int date) {
    string s = to_string(date);
    if (s[0] == s[2] && s[0] == s[5] && s[0] == s[7] && 
        s[1] == s[3] && s[1] == s[4] && s[1] == s[6])
        return true;
    return false;
}
 
signed main() {
    cin >> date;
    int y = date / 10000, m = date / 100 % 100, d = date % 100;
 
    for (int i = y;; i++) {
        if (i % 400 == 0 || (i % 100 != 0 && i % 4 == 0))
            month[2] = 29;
        else
            month[2] = 28;
 
        int j = (i == y) ? m : 1;
        for (; j <= 12; j++) {
            int k = (i == y && j == m) ? d + 1 : 1;
            for (; k <= month[j]; k++) {
                int new_date = i * 10000 + j * 100 + k;
                if (check1(new_date) && ans1 == 0)
                    ans1 = new_date;
                if (check2(new_date)) {
                    cout << ans1 << '\n' << new_date << '\n';
                    return 0; // 当找到了 ABABBABA 型回文日期时，结束程序
                }
            }
        }
    }
 
    return 0;
}

枚举年份：

#include <bits/stdc++.h>
using namespace std;
 
// month[1~12]分别记录1~12月每月的天数
int date, month[13] = {-1, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
 
// 判断日期是否合法
string check(int date) {
    string s = to_string(date), t = to_string(date); // 将整型转换为字符串
    reverse(t.begin(), t.end()); // 翻转
    s += t; // 将s翻转后再与s拼接，拼接后的字符串一定会是个回文串
    int y = stoi(s.substr(0, 4)), m = stoi(s.substr(4, 2)), d = stoi(s.substr(6, 2));
 
    if (y % 400 == 0 || (y % 4 == 0 && y % 100 != 0)) 
        month[2] = 29;
    else 
        month[2] = 28;
 
    if (m < 1 || m > 12 || d < 1 || d > month[m]) 
        return "-1";
    return s;
}
 
// 判断日期是否是ABABBABA型回文
string check2(int date) {
    string s = to_string(date), t = to_string(date);
    reverse(t.begin(), t.end()); // 翻转
    s += t;
    if (s[0] == s[2] && s[0] == s[5] && s[0] == s[7] && 
        s[1] == s[3] && s[1] == s[4] && s[1] == s[6]) 
        return s;
    return "-1";
}
 
signed main() {
    cin >> date;
    string ans1 = "";
 
    for (int i = date / 10000;; i++) {
        // 注意：date（输入的日期）不能作为答案
        if (check(i) == "-1" || check(i) == to_string(date)) continue;
        if (ans1 == "") ans1 = check(i);
        if (check2(i) != "-1") {
            cout << ans1 << '\n' << check2(i) << '\n';
            return 0;
        }
    }
 
    return 0;
}

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