//	C++年月日判断初步代码
#include <iostream>

using namespace std;

class Data
{
	int year;
	int month;
	int day;
public:
	//判断日期是否有效
	bool data_check(int _year,int _month,int _day)
	{
		year = _year;
		month =  _month;
		day = _day;

		if(0 > year || 2018 < year) return false;
		if(0 > month || 12 < month) return false;
		if(0 > day || 31 < day) return false;
		return true;
	}
	//判断是否是闰年
	bool leap_year(int _year)
	{
		year = _year;
		if(year%400 == 0 || (year%4 == 0 &&year%100 != 0))//能被4,400整除除去被100整除的都是闰年
		{
			return true;
		}
		else 
		{
			return false;
		}
	}
	//下一天的日期
	void nextday(int _year,int _month ,int _day)
	{
		
		year = _year;
		month = _month;
		day = _day;
		switch(month)
		{
			case 1:
				if(31 == day)
				{
					month++;
					day = 1;
					break;
				}
			case 2:
				if((28 == day &&!leap_year(year))||(29 == day &&leap_year(year)))
				{
					month++;
					day = 1;
					break;
				}
			case 3:
				if(31 == day)
				{
					month++;
					day = 1;
					break;
				}
			case 4:
				if(30 == day)
				{
					month++;
					day = 1;
					break;
				}
			case 5:
				if(31 == day)
				{
					month++;
					day =1;
					break;
				}
			case 6:
				if(30 == day)
				{
					month++;
					day =1;
					break;
				}
			case 7:
				if(31 == day)
				{
					month++;
					day = 1;
					break;
				}
			case 8:
				if(31 == day)
				{
					month++;
					day =1;
					break;
				}
			case 9:
				if(30 == day)
				{
					month++;
					day = 1;
					break;
				}
			case 10:
				if(31 == day)
				{
					month++;
					day = 1;
					break;
				}
			case 11:
				if(30 == day)
				{
					month++;
					day = 1;
					break;
				}
			case 12:
				if(31 == day)
				{
					year++;
					month = 1;
					day = 1;
					break;
				}
		}
		cout << year << "." << month << "." << day << endl;
	}
	//列出下n天或者前n天的日期
	void nextday(int _year,int _month,int _day,int n)
	{
		year = _year;
		month = _month;
		day = _day;
		if(0 < n)
		{
			
			for(n =n+1;n>0;n--)
			{
				switch(month)
				{
					case 1:
						if(31 == day)
						{
							month++;
							day = 1;
							break;
						}else {
							day++;
							break;
						}
					case 2:
						if((28 == day &&!leap_year(year))||(29 == day &&leap_year(year)))
						{
							month++;
							day = 1;
							break;
						}else {
							day++;
							break;
						}
					case 3:
						if(31 == day)
						{
							month++;
							day = 1;
							break;
						}else {
							day++;
							break;
						}
					case 4:
						if(30 == day)
						{
							month++;
							day = 1;
							break;
						}else {
							day++;
							break;
						}
					case 5:
						if(31 == day)
						{
							month++;
							day =1;
							break;
						}
					case 6:
						if(30 == day)
						{
							month++;
							day =1;
							break;
						}else {
							day++;
							break;
						}
					case 7:
						if(31 == day)
						{
							month++;
							day = 1;
							break;
						}else day++;
					case 8:
						if(31 == day)
						{
							month++;
							day =1;
							break;
						}else {
							day++;
							break;
						}
					case 9:
						if(30 == day)
						{
							month++;
							day = 1;
							break;
						}else {
							day++;
							break;
						}
					case 10:
						if(31 == day)
						{
							month++;
							day = 1;
							break;
						}else {
							day++;
							break;
						}
					case 11:
						if(30 == day)
						{
							month++;
							day = 1;
							break;
						}else {
							day++;
							break;
						}
					case 12:
						if(31 == day)
						{
							year++;
							month = 1;
							day = 1;
							break;
						}else {
							day++;
							break;
						}
				}
			cout << year << "." << month << "." << day << endl;
			}
		}
		if(0 > n)
		{
			
			for(n =n-1;n<0;n++)
			{
				switch(month)
				{
					case 1:
						if(1 == day)
						{
							year--;							
							month =12;
							day = 31;
							break;
						}else {
							day--;
							break;
						}
					case 2:
						if(1 == day)
						{
							month--;
							day = 31;
							break;
						}else {
							day--;
							break;
						}
					case 3:
						if(1 == day)
						{
							month--;
							if(leap_year(year))
							{
								day = 29;
							}else{
								day = 28;
							}
							break;
						}else {
							day--;
							break;
						}
					case 4:
						if(1 == day)
						{
							month--;
							day = 31;
							break;
						}else {
							day--;
							break;
						}
					case 5:
						if(1 == day)
						{
							month--;
							day =30;
							break;
						}else {
							day--;
							break;
						}
					case 6:
						if(1 == day)
						{
							month--;
							day =31;
							break;
						}else {
							day--;
							break;
						}
					case 7:
						if(1 == day)
						{
							month--;
							day = 30;
							break;
						}else {
							day--;
							break;
						}
					case 8:
						if(1 == day)
						{
							month--;
							day =31;
							break;
						}else {
							day--;
							break;
						}
					case 9:
						if(1 == day)
						{
							month--;
							day = 31;
							break;
						}else {
							day--;
							break;
						}
					case 10:
						if(1 == day)
						{
							month--;
							day = 30;
							break;
						}else {
							day--;
							break;
						}
					case 11:
						if(1 == day)
						{
							month--;
							day = 31;
							break;
						}else {
							day--;
							break;
						}
					case 12:
						if(1 == day)
						{
							month--;
							day = 31;
							break;
						}else {
							day--;
							break;
						}
				}
			cout << year << "." << month << "." << day << endl;
			}
		}
	}
	

};
int main()
{
	Data *data = new Data;
	int year = 1997;
	int month = 2;
	int day = 28;
	int n = -137;
	cout << data->data_check(year,month,day) << endl;
	cout << data->leap_year(year) << endl;
	data-> nextday(year,month,day);
	data-> nextday(year,month,day,n);
	
	
}

C++简单年月日的功能实现的更多相关文章

  1. ASP.NET MVC 学习4、Controller中添加SearchIndex页面,实现简单的查询功能

    参考:http://www.asp.net/mvc/tutorials/mvc-4/getting-started-with-aspnet-mvc4/examining-the-edit-method ...

  2. Web---创建Servlet的3种方式、简单的用户注册功能

    说明: 创建Servlet的方式,在上篇博客中,已经用了方式1(实现Servlet接口),接下来本节讲的是另外2种方式. 上篇博客地址:http://blog.csdn.net/qq_26525215 ...

  3. js+html+css简单的互动功能页面(2015知道几乎尖笔试题)http://v.youku.com/v_show/id_XMTI0ODQ5NTAyOA==.html?from=y1.7-1.2

    js+html+css实现简单页面交互功能(2015知乎前端笔试题) http://v.youku.com/v_show/id_XMTI0ODQ5NTAyOA==.html? from=y1.7-1. ...

  4. Spring 学习——基于Spring WebSocket 和STOMP实现简单的聊天功能

    本篇主要讲解如何使用Spring websocket 和STOMP搭建一个简单的聊天功能项目,里面使用到的技术,如websocket和STOMP等会简单介绍,不会太深,如果对相关介绍不是很了解的,请自 ...

  5. Django文件上传三种方式以及简单预览功能

    主要内容: 一.文件长传的三种方式 二.简单预览功能实现 一.form表单上传 1.页面代码 <!DOCTYPE html> <html lang="en"> ...

  6. 运用socket实现简单的ssh功能

    在python socket知识点中已经对socket进行了初步的了解,那现在就使用这些知识来实现一个简单的ssh(Secure Shell)功能. 首先同样是建立两个端(服务器端和客户端) 需求是: ...

  7. Jenkins实现简单的CI功能

    步骤一:安装JDK.Tomcat,小儿科的东西不在此详细描述 步骤二:下载安装Jenkins下载链接:https://jenkins.io/download/ 步骤三:将下载的jenkins.war部 ...

  8. jQuery实现简单前端搜索功能

    <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...

  9. Unity UGUI 实现简单拖拽功能

    说到拖拽,那必然离不开坐标,UGUI 的坐标有点不一样,它有两种坐标,一种是屏幕坐标,还有一种就是 UI 在Canvas内的坐标(暂时叫做ugui坐标),这两个坐标是不一样的,所以拖拽就需要转换. 因 ...

随机推荐

  1. python之文件路径截取 & endswith()

    文件路径截取: >>> import os >>> path = '/etc/singfor/passwd/sunny/test.log' >>> ...

  2. JSP && Servlet | 错误统一处理

    对404错误和500错误处理: 在WebContent文件下新建404.jsp 和 500.jsp 显示错误时弹出的信息 <%@ page language="java" c ...

  3. 管理现有数据库-web系统

    1 需求 现有的业务数据需要经常被展示,所以选择django作为展示工具.只需要使用django自带的admin app,然后对现有数据库进行建模就可以搞定. 2 代码 settings: DATAB ...

  4. 洛谷 P4549 【模板】裴蜀定理

    https://www.luogu.org/problemnew/show/P4549 (1)证明方程ax+by=gcd(a,b)(a,b为常数;a>0,b>0;a,b,x,y为整数)有解 ...

  5. centOS6.5 usr/src/kernels下为空

    用uname -r查看内核版本为 2.6.32-431.el6.x86_64 usr/src/kernels下为空 需要执行两个安装 yum install kernel-headers yum in ...

  6. python入门之实例-验证码

    需求: 随机生成6位的验证码,要求有字母和数字 import random temp = "" for i in range(6): j = random.randrange(0, ...

  7. POJ 1830 开关问题 高斯消元,自由变量个数

    http://poj.org/problem?id=1830 如果开关s1操作一次,则会有s1(记住自己也会变).和s1连接的开关都会做一次操作. 那么设矩阵a[i][j]表示按下了开关j,开关i会被 ...

  8. 将GitLab上面的代码克隆到本地

    1.安装GitLab客户端 2.去GitLab服务端找项目路径 3.去GitLab客户端去克隆代码 右键-->git Clone 4.最后结果

  9. PeopleSoft FSCM Production Support 案例分析之一重大紧急事故发生时的应对策略

    案例背景: 今天一大早用户打电话来讲昨天上传的银行的forex payment return file好像没有被处理到,我一听就觉得纳闷,因为昨天晚上operator也没有给我打电话啊(如果有job ...

  10. Objective-C Loops

    There may be a situation, when you need to execute a block of code several number of times. In gener ...