C++给定一个日期, 计算n天之后的日期-蒲公英云

C++给定一个日期, 计算n天之后的日期

设计一个日期类，包含以下功能：
1、只能通过传入年月日初始化。
2、可以加上一个数字n，返回一个该日期后推n天之后的日期。

值得注意的是, 闰年并不是4年一闰, 我们通常习惯理解为4年一闰
闰年判断条件 (year % 400 == 0 || (year % 4 == 0 && year % 100)), 可以看到, 当年份是4的倍数, 但可以被100整除, 且不能被400整除时, 并不是闰年, 例如2096年是闰年, 按习惯2100年是闰年, 但2100年不能满足判断条件,不是闰年,下一个闰年是2104, 相隔8年, 还可以发现 2100,2200,2300,2500,2600,2700,2900等等都不是闰年, 遇到这种年份前后闰年相隔8年

Date.h

#pragma once
#include<iostream>
using namespace std;
class Date {
    static int m_s_month[12];
    int m_year;
    int m_month;
    int m_day;
public:
    Date(int y = 0, int m = 0, int d = 0) {
        m_year = y;
        m_month = m;
        m_day = d;
    }
    friend ostream& operator<<(ostream& os, Date& a);
    friend istream& operator>>(istream& os, Date& a);
    static int getMonthDay(int y, size_t n);//返回y年n月的天数
    bool LeapYear(int year)const;
    int countLeapYear(int year)const;
    Date operator+(size_t n)const;
};

Date.cpp

#include "Date.h"
int Date::m_s_month[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
int Date::getMonthDay(int y, size_t n) {
    return m_s_month[n - 1] + (n == 2 && (y % 400 == 0 || (y % 4 == 0 && y % 100)));
}
bool Date::LeapYear(int y)const {
    return y % 400 == 0 || (y % 4 == 0 && y % 100 != 0);
}
ostream& operator<<(ostream& os, Date& a) {
    os << a.m_year << "年" << a.m_month << "月" << a.m_day << "日";
    return os;
}
istream& operator>>(istream& os, Date& t) {
    printf("请输入年份\n");
    while (1) {
        os >> t.m_year;
        if (t.m_year >= 0) {
            break;
        }
        cout << "年份必须非负,请重新输入年份\n";
    }
    cout << "请输入月份\n";
    while (1) {
        os >> t.m_month;
        if (t.m_month < 1) {
            cout << "月份必须大于0, 请重新输入\n";
            continue;
        }
        if (t.m_month > 12) {
            cout << "月份必须小于等于12, 请重新输入\n";
            continue;
        }
        break;
    }
    cout << "请输入日\n";
    while (1) {
        os >> t.m_day;
        if (t.m_day < 1) {
            cout << "请输入大于0的天数\n";
            continue;
        }
        if (t.m_day > Date::getMonthDay(t.m_year, t.m_month)) {
            cout << "你输入的号数大于本月天数" << Date::getMonthDay(t.m_year, t.m_month) << "请重新输入\n";
            continue;
        }
        break;
    }
    return os;
}
int Date::countLeapYear(int year)const {
    if (m_year == year) {
        return LeapYear(year);
    }
    int begin = m_year;
    int end = year - 1;
    if (m_year > year) {
        begin = year;
        end = m_year - 1;
    }
    while (!(LeapYear(begin))) {
        ++begin;
    }
    while (!(LeapYear(end))) {
        --end;
    }
    int tmp = begin;
    int count = 0;
    if (begin <= end) {
        while (tmp <= end && tmp % 100) {
            ++tmp;;
        }
        for (; tmp <= end; !LeapYear(tmp) ? ++count : count, tmp += 100);
        return (end - begin) / 4 + 1 - count;
    }
    return 0;
}
Date Date::operator+(size_t n)const {
    Date tmp = *this;
    size_t count = 0;
    for (int i = 1; i < (int)tmp.m_month; ++i) {
        count += getMonthDay(tmp.m_year, i);
    }
    count += tmp.m_day;
    if (n >= (365 + LeapYear(tmp.m_year) - count)) {
        n -= (365 + LeapYear(tmp.m_year) - count);
        tmp.m_month = 12;
        tmp.m_day = 31;
        while ((int)n >= (365 + LeapYear(tmp.m_year + 1))) {
            n -= (365 + LeapYear(++tmp.m_year));
        }
    }
    for (; n > 0; --n) {
        if (tmp.m_day == getMonthDay(tmp.m_year, tmp.m_month)) {
            tmp.m_day = 1;
            tmp.m_month == 12 ? tmp.m_month = 1, ++tmp.m_year : ++tmp.m_month;
        }
        else {
            ++tmp.m_day;
        }
    }
    return tmp;
}

main.cpp

#include"Date.h"
int main() {
    int n;
    Date a, b;
    cout << "请输入日期\n";
    cin >> a;
    cout << "你想知道多少天后的日期,请输入\n";
    cin >> n;
    b = a + n;
    cout << a << endl;
    cout << "在" << n << "天后是";
    cout << b << endl;
    system("pause");
    return 0;
}