前言

在做题中遇到了几次的日期问题之后，与大家分享一下相关日期问题常用的模板，如果喜欢可以点个赞！

模板

#include <iostream>
using namespace std;

int months[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

// 判断日期的合法性
bool check_valid(int date) //形如20210305
{
    int year = date / 10000;
    int month = date % 10000 / 100;
    int day = date % 100;

    if (month <= 0 || month >= 13) return false;
    if (day == 0 || month != 2 && day > months[month]) return false;

    if (month == 2)
    {
        int leap = (year % 4 == 0 && year % 100 != 0 || year % 400 == 0);
        if (day > 28 + leap) return false;
    }

    return true;
}

// 得到某年某月的天数
int get(int year, int month) 
{
    if (month != 2) return months[month];
    else
    {
        // 2月
        int leap = (year % 4 == 0 && year % 100 != 0 || year % 400 == 0);
        return 28 + leap;
    }
}

// 判断两个日期之间有多少个回文日期
int get(int date1, int date2, int k) //k使函数签名不同从而编译通过
{
    int ans = 0;
    for (int i = 1000; i < 10000; i++)
    {
        int date = i, x = i;
        for (int j = 0; j < 4; j++) date = date * 10 + x % 10, x /= 10; //根据年份构造出回文日期
        if (date1 <= date && date <= date2 && check_valid(date)) ans++; 
    }

    return ans;
}

// 给定年月日，经过n天后对应的日期
void pass(int y, int m, int d, int n)
{
    while (n--)
    {
        d++;
        if (d > get(y, m)) m++, d = 1;
        if (m > 12) y++, m = 1;
    }
    printf("%d-%02d-%02d\n", y, m, d);
}

int main()
{
    return 0;
}

参考练习题（附上代码）

466. 回文日期：https://www.acwing.com/problem/content/468/

#include <iostream>
using namespace std;

int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

bool check_valid(int date)    //1234 5678
{
    int year = date / 10000;
    int month = date % 10000 / 100;
    int day = date % 100;

    if (month <= 0 || month >= 13) return false;
    if (day == 0 || month != 2 && day > days[month]) return false;

    if (month == 2)
    {
        int leap = (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
        if (day > 28 + leap) return false;
    }

    return true;
}

int main()
{
    int date1, date2;
    cin >> date1 >> date2;

    int ans = 0;
    for (int i = 1000; i < 10000; i++)
    {
        int date = i, x = i;
        for (int j = 0; j < 4; j++) date = date * 10 + x % 10, x /= 10;
        if (date1 <= date && date <= date2 && check_valid(date)) ans++;
    }

    cout << ans << endl;
    return 0;
}

1229. 日期问题：https://www.acwing.com/problem/content/1231/

#include <iostream>
using namespace std;

int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

bool check_valid(int year, int month, int day)
{
    if (month <= 0 || month >= 13) return false;
    if (day == 0 || month != 2 && day > days[month]) return false;

    if (month == 2)
    {
        int leap = (year % 400 == 0 || year % 4 == 0 && year % 100 != 0);
        if (day > 28 + leap) return false;
    }

    return true;
}

int main()
{
    int a, b, c;
    scanf("%d/%d/%d", &a, &b, &c);

    for (int i = 19600101; i <= 20591231; i++)
    {
        int year = i / 10000, month = i % 10000 / 100, day = i % 100;
        if (check_valid(year, month, day))
        {
            if (year % 100 == a && month == b && day == c ||
                month == a && day == b && year % 100 == c ||
                day == a && month == b && year % 100 == c)
                    printf("%d-%02d-%02d\n", year, month, day);
        }
    }

    return 0;
}

2867. 回文日期：https://www.acwing.com/problem/content/2870/

#include <iostream>
using namespace std;

int months[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

// 判断日期的合法性
bool check_valid(int date) //形如20210305
{
    int year = date / 10000;
    int month = date % 10000 / 100;
    int day = date % 100;

    if (month <= 0 || month >= 13) return false;
    if (day == 0 || month != 2 && day > months[month]) return false;

    if (month == 2)
    {
        int leap = (year % 4 == 0 && year % 100 != 0 || year % 400 == 0);
        if (day > 28 + leap) return false;
    }

    return true;
}

// 判断date是否满足ABABBABA形式
bool check(int date)
{
    int year = date / 10000;
    int month = date % 10000 / 100;
    int day = date % 100;


    if (month % 10 == day / 10) return false; //A==B return false
    if ((month % 10 == day % 10) && (month / 10 == day / 10)) return true;

    return false;
}

bool st1, st2;
int ans1, ans2;

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

    int year = date / 10000;
    while (true)
    {
        int new_date = year, x = year;
        for (int i = 0; i < 4; i++) new_date = new_date * 10 + x % 10, x /= 10; //构造回文日期

        if (new_date == date)  //构造出来的是同一天
        {
            year++;
            continue;    
        }

        if (check_valid(new_date))
        {
            if (!st1) st1 = true, ans1 = new_date;
            if (!st2 && check(new_date)) st2 = true, ans2 = new_date;

        }

        year ++;
        if (st1 && st2) break; //找到两个解则退出
    }

    printf("%d\n%d", ans1, ans2);
    return 0;
}

3218. 日期计算：https://www.acwing.com/problem/content/3221/

#include <iostream>
using namespace std;

int months[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

// 得到某年某月的天数
int get(int year, int month)
{
    if (month != 2) return months[month];
    else
    {
        // 2月
        int leap = (year % 4 == 0 && year % 100 != 0 || year % 400 == 0);
        return 28 + leap;
    }
}

// 给定年月日，经过n天后对应的日期
void pass(int y, int m, int d, int n)
{
    while (n--)
    {
        d++;
        if (d > get(y, m)) m++, d = 1;
        if (m > 12) y++, m = 1;
    }
    printf("%d\n%d\n", m, d);
}

int main()
{
    int y, n;
    cin >> y >> n;
    int m = 1, d = 0;
    pass(y, 1, d, n);
    return 0;
}