C++给定一个日期, 计算n天之后的日期

不念不忘少年蓝@ 2024-04-17 23:20 75阅读 0赞

> 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;
    }

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