类似题 - 466. 回文日期

写法2，先生成回文日期，再判断合不合法【常用技巧！】

#include <iostream>
#include <algorithm>

using namespace std;

int x;
int days[13] = {0,31,29,31,30,31,30,31,31,30,31,30,31};

bool valid(int x)
{
    int year = x / 10000;
    int month = x % 10000 / 100;
    int day = x % 100;

    if (month == 0 || month > 12) return false;
    if (day == 0 || month != 2 && day > days[month]) return false;

    if (month == 2)
    {
        bool leap = year % 100 && year % 4 ==0 || year % 400 == 0; //判断闰年
        if (day > 28 + leap) return false;
    }

    return true;
}

bool check_ABAB(int x)
{

    string s = to_string(x);
    return s[0] == s[2] && s[1] == s[3] && s[0] != s[1];
}

int main()
{
    int x;
    cin >> x;

    bool f = 1;
    for(int i = 1000;i  ;i ++) // 根据年份构造回文日期
    {
        int data = i, t = i;
        for(int j = 0;j < 4;j ++ ) data = data * 10 + t % 10, t /= 10; // 生成回文日期

        if(data > x && valid(data) && f){
            cout << data <<endl;
            f = 0;
        }
        if(data > x && valid(data) && check_ABAB(data)){
            cout << data;
            break;
        }
    }

    return 0;
}

写法1，暴力枚举数字

#include <iostream>
#include <algorithm>

using namespace std;

int x;
int days[13] = {0,31,29,31,30,31,30,31,31,30,31,30,31};

bool valid(int x)
{
    int year = x / 10000;
    int month = x % 10000 / 100;
    int day = x % 100;

    if (month == 0 || month > 12) return false;
    if (day == 0 || month != 2 && day > days[month]) return false;

    if (month == 2)
    {
        bool leap = year % 100 && year % 4 ==0 || year % 400 == 0; //判断闰年
        if (day > 28 + leap) return false;
    }

    return true;
}

bool check(int x)
{
    int t = 0;
    int xx = x;
    while(x){
        t = t * 10 + x % 10;
        x /= 10;
    }
    return xx == t;
}

bool check2(int x)
{
    if(!check(x)) return false;

    if( x % 10 == (x / 10) % 10) return false;

    string s = to_string(x);
    return s[0] == s[2] && s[1] == s[3];

}

int main()
{
    int x;
    cin >> x;

    bool f = 1,f2 = 1;
    for(int i = x+ 1; ;i ++){ // 是下一个
        if(valid(i)) // 合法日期
        {
            if(check(i) && f){
                cout << i << endl;
                f = 0;
            }
            if(check2(i) && f2){
                cout << i << endl;
                f2 = 0;
            }

            if(!f && !f2) break;
        }

    }

    return 0;
}