给定任意日期得出星期几代码,C代码库,德仔网

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【C】给定任意日期得出星期几
作者:Dezai.CN / 发布于2013/2/22/ 671
任意日期得出星期几.cpp : 定义控制台应用程序的入口点

	// 任意日期得出星期几.cpp : 定义控制台应用程序的入口点。
	//

	#include "stdafx.h"
	#include<iostream>
	#include<string>
	#include<vector>
	//#include<algorithm>
	using namespace std;

	int a[]={2013,2,17};
	vector<int> vec(a,a+3);
	class time
	{
	public:
	time(){}
	time(const vector<int>& v):vec(v){}
	friend istream& operator>>(istream&,time&);
	friend ostream& operator<<(ostream&,const time&);
	friend long operator-(time&,time&);//计算两个日期的相隔天数
	bool test();//测试输入的日期是否合法
	time* preday();//返回给定日期的前一天
	bool compare(const time&);//与2013-2-17比较大小
	private:
	vector<int> vec;
	};
	long operator-(time& t1,time&t2)////两个具体日期的减法
	{
	long sum=0;
	time tt1=t1,tt2=t2;/////tt1存放的日期大于tt2存放的日期
	if(!t1.compare(t2))//判断两个日期的大小
	{
	 tt1=t2;
	 tt2=t1;
	}
	if(tt1.vec[0]>tt2.vec[0])//判断两个年份大小，大于则分别计算该两个年份中的要求天数和两年份之间的年份的天数
	 {
	switch(tt2.vec[1])
	{
	case 1:
	 sum+=365-tt2.vec[2];
	 break;
	case 2:
	 sum+=365-31-tt2.vec[2];
	 break;
	case 3:
	 sum+=365-31-28-tt2.vec[2];
	 break;
	case 4:
	 sum+=365-31-28-31-tt2.vec[2];
	 break;
	case 5:
	 sum+=365-31-28-31-30-tt2.vec[2];
	 break;
	case 6:
	 sum+=365-31-28-31-30-31-tt2.vec[2];
	 break;
	case 7:
	 sum+=365-31-28-31-30-31-30-tt2.vec[2];
	 break;
	case 8:
	 sum+=365-31-28-31-30-31-30-31-tt2.vec[2];
	 break;
	case 9:
	 sum+=365-31-28-31-30-31-30-31-31-tt2.vec[2];
	 break;
	case 10:
	 sum+=365-31-28-31-30-31-30-31-31-30-tt2.vec[2];
	 break;
	case 11:
	 sum+=365-31-28-31-30-31-30-31-31-30-31-tt2.vec[2];
	 break;
	case 12:
	 sum+=365-31-28-31-30-31-30-31-31-30-31-30-tt2.vec[2];
	 break;
	}
	if((tt2.vec[0]%400==0||(tt2.vec[0]%100!=0&&tt2.vec[0]%4==0))&&(tt2.vec[1]==1||tt2.vec[1]==2))
	 sum+=1;//闰年并满足(tt2.vec[1]==1||tt2.vec[1]==2)则总天数加一
	 for(long i=1;i<tt1.vec[0]-tt2.vec[0];i++)
	 {
	 if((tt2.vec[0]+i)%400==0||((tt2.vec[0]+i)%100!=0&&(tt2.vec[0]+i)%4==0))
	 sum+=366;
	 else
	 sum+=365;
	 }
	 switch(tt1.vec[1])//计算大的日期中已过的天数
	 {
	case 1:
	 sum+=tt1.vec[2];
	 break;
	case 2:
	 sum+=31+tt1.vec[2];
	 break;
	case 3:
	 sum+=31+28+tt1.vec[2];
	 break;
	case 4:
	 sum+=31+28+31+tt1.vec[2];
	 break;
	case 5:
	 sum+=31+28+31+30+tt1.vec[2];
	 break;
	case 6:
	 sum+=31+28+31+30+31+tt1.vec[2];
	 break;
	case 7:
	 sum+=31+28+31+30+31+30+tt1.vec[2];
	 break;
	case 8:
	 sum+=31+28+31+30+31+30+31+tt1.vec[2];
	 break;
	case 9:
	 sum+=31+28+31+30+31+30+31+31+tt1.vec[2];
	 break;
	case 10:
	 sum+=31+28+31+30+31+30+31+31+30+tt1.vec[2];
	 break;
	case 11:
	 sum+=31+28+31+30+31+30+31+31+30+31+tt1.vec[2];
	 break;
	case 12:
	 sum+=31+28+31+30+31+30+31+31+30+31+30+tt1.vec[2];
	 break;
	 }
	 if((tt1.vec[0]%400==0||(tt1.vec[0]%100!=0&&tt1.vec[0]%4==0))&&(tt1.vec[1]!=1&&tt1.vec[1]!=2))
	 sum+=1;
	 }
	else//两个日期的年份即vec[0]相等时
	{
	 if(tt2.vec[1]==tt1.vec[1])
	 sum+=tt1.vec[2]-tt2.vec[2];
	 else
	 {
	 switch(tt2.vec[1])
	 {
	 case 1:
	 switch(tt1.vec[1])
	 {
	 case 2:
	 sum+=31-tt2.vec[2]+tt1.vec[2];
	 break;
	 case 3:
	 sum+=31-tt2.vec[2]+28+tt1.vec[2];
	 break;
	 case 4:
	 sum+=31-tt2.vec[2]+28+31+tt1.vec[2];
	 break;
	 case 5:
	 sum+=31-tt2.vec[2]+28+31+30+tt1.vec[2];
	 break;
	 case 6:
	 sum+=31-tt2.vec[2]+28+31+30+31+tt1.vec[2];
	 break;
	 case 7:
	 sum+=31-tt2.vec[2]+28+31+30+31+30+tt1.vec[2];
	 break;
	 case 8:
	 sum+=31-tt2.vec[2]+28+31+30+31+30+31+tt1.vec[2];
	 break;
	 case 9:
	 sum+=31-tt2.vec[2]+28+31+30+31+30+31+31+tt1.vec[2];
	 break;
	 case 10:
	 sum+=31-tt2.vec[2]+28+31+30+31+30+31+31+30+tt1.vec[2];
	 break;
	 case 11:
	 sum+=31-tt2.vec[2]+28+31+30+31+30+31+31+30+31+tt1.vec[2];
	 break;
	 case 12:
	 sum+=31-tt2.vec[2]+28+31+30+31+30+31+31+30+31+30+tt1.vec[2];
	 break;
	 }
	 break;
	 case 2:
	 switch(tt1.vec[1])
	 {
	 case 3:
	 sum+=28-tt2.vec[2]+tt1.vec[2];
	 break;
	 case 4:
	 sum+=28-tt2.vec[2]+31+tt1.vec[2];
	 break;
	 case 5:
	 sum+=28-tt2.vec[2]+31+30+tt1.vec[2];
	 break;
	 case 6:
	 sum+=28-tt2.vec[2]+31+30+31+tt1.vec[2];
	 break;
	 case 7:
	 sum+=28-tt2.vec[2]+31+30+31+30+tt1.vec[2];
	 break;
	 case 8:
	 sum+=28-tt2.vec[2]+31+30+31+30+31+tt1.vec[2];
	 break;
	 case 9:
	 sum+=28-tt2.vec[2]+31+30+31+30+31+31+tt1.vec[2];
	 break;
	 case 10:
	 sum+=28-tt2.vec[2]+31+30+31+30+31+31+30+tt1.vec[2];
	 break;
	 case 11:
	 sum+=28-tt2.vec[2]+31+30+31+30+31+31+30+31+tt1.vec[2];
	 break;
	 case 12:
	 sum+=28-tt2.vec[2]+31+30+31+30+31+31+30+31+30+tt1.vec[2];
	 break;
	 }
	 break;
	 case 3:
	 switch(tt1.vec[1])
	 {
	 case 4:
	 sum+=31-tt2.vec[2]+tt1.vec[2];
	 break;
	 case 5:
	 sum+=31-tt2.vec[2]+30+tt1.vec[2];
	 break;
	 case 6:
	 sum+=31-tt2.vec[2]+30+31+tt1.vec[2];
	 break;
	 case 7:
	 sum+=31-tt2.vec[2]+30+31+30+tt1.vec[2];
	 break;
	 case 8:
	 sum+=31-tt2.vec[2]+30+31+30+31+tt1.vec[2];
	 break;
	 case 9:
	 sum+=31-tt2.vec[2]+30+31+30+31+31+tt1.vec[2];
	 break;
	 case 10:
	 sum+=31-tt2.vec[2]+30+31+30+31+31+30+tt1.vec[2];
	 break;
	 case 11:
	 sum+=31-tt2.vec[2]+30+31+30+31+31+30+31+tt1.vec[2];
	 break;
	 case 12:
	 sum+=31-tt2.vec[2]+30+31+30+31+31+30+31+30+tt1.vec[2];
	 break;
	 }
	 break;
	 case 4:
	 switch(tt1.vec[1])
	 {
	 case 5:
	 sum+=30-tt2.vec[2]+tt1.vec[2];
	 break;
	 case 6:
	 sum+=30-tt2.vec[2]+31+tt1.vec[2];
	 break;
	 case 7:
	 sum+=30-tt2.vec[2]+31+30+tt1.vec[2];
	 break;
	 case 8:
	 sum+=30-tt2.vec[2]+31+30+31+tt1.vec[2];
	 break;
	 case 9:
	 sum+=30-tt2.vec[2]+31+30+31+31+tt1.vec[2];
	 break;
	 case 10:
	 sum+=30-tt2.vec[2]+31+30+31+31+30+tt1.vec[2];
	 break;
	 case 11:
	 sum+=30-tt2.vec[2]+31+30+31+31+30+31+tt1.vec[2];
	 break;
	 case 12:
	 sum+=30-tt2.vec[2]+31+30+31+31+30+31+30+tt1.vec[2];
	 break;
	 }
	 break;
	 case 5:
	 switch(tt1.vec[1])
	 {
	 case 6:
	 sum+=31-tt2.vec[2]+tt1.vec[2];
	 break;
	 case 7:
	 sum+=31-tt2.vec[2]+30+tt1.vec[2];
	 break;
	 case 8:
	 sum+=31-tt2.vec[2]+30+31+tt1.vec[2];
	 break;
	 case 9:
	 sum+=31-tt2.vec[2]+30+31+31+tt1.vec[2];
	 break;
	 case 10:
	 sum+=31-tt2.vec[2]+30+31+31+30+tt1.vec[2];
	 break;
	 case 11:
	 sum+=31-tt2.vec[2]+30+31+31+30+31+tt1.vec[2];
	 break;
	 case 12:
	 sum+=31-tt2.vec[2]+30+31+31+30+31+30+tt1.vec[2];
	 break;
	 }
	 break;
	 case 6:
	 switch(tt1.vec[1])
	 {
	 case 7:
	 sum+=30-tt2.vec[2]+tt1.vec[2];
	 break;
	 case 8:
	 sum+=30-tt2.vec[2]+31+tt1.vec[2];
	 break;
	 case 9:
	 sum+=30-tt2.vec[2]+31+31+tt1.vec[2];
	 break;
	 case 10:
	 sum+=30-tt2.vec[2]+31+31+30+tt1.vec[2];
	 break;
	 case 11:
	 sum+=30-tt2.vec[2]+31+31+30+31+tt1.vec[2];
	 break;
	 case 12:
	 sum+=30-tt2.vec[2]+31+31+30+31+30+tt1.vec[2];
	 break;
	 }
	 break;
	 case 7:
	 switch(tt1.vec[1])
	 {
	 case 8:
	 sum+=31-tt2.vec[2]+tt1.vec[2];
	 break;
	 case 9:
	 sum+=31-tt2.vec[2]+31+tt1.vec[2];
	 break;
	 case 10:
	 sum+=31-tt2.vec[2]+31+30+tt1.vec[2];
	 break;
	 case 11:
	 sum+=31-tt2.vec[2]+31+30+31+tt1.vec[2];
	 break;
	 case 12:
	 sum+=31-tt2.vec[2]+31+30+31+30+tt1.vec[2];
	 break;
	 }
	 break;
	 case 8:
	 switch(tt1.vec[1])
	 {
	 case 9:
	 sum+=31-tt2.vec[2]+tt1.vec[2];
	 break;
	 case 10:
	 sum+=31-tt2.vec[2]+30+tt1.vec[2];
	 break;
	 case 11:
	 sum+=31-tt2.vec[2]+30+31+tt1.vec[2];
	 break;
	 case 12:
	 sum+=31-tt2.vec[2]+30+31+30+tt1.vec[2];
	 break;
	 }
	 break;
	 case 9:
	 switch(tt1.vec[1])
	 {
	 case 10:
	 sum+=30-tt2.vec[2]+tt1.vec[2];
	 break;
	 case 11:
	 sum+=30-tt2.vec[2]+31+tt1.vec[2];
	 break;
	 case 12:
	 sum+=30-tt2.vec[2]+31+30+tt1.vec[2];
	 break;
	 }
	 break;
	 case 10:
	 switch(tt1.vec[1])
	 {
	 case 11:
	 sum+=31-tt2.vec[2]+tt1.vec[2];
	 break;
	 case 12:
	 sum+=31-tt2.vec[2]+30+tt1.vec[2];
	 break;
	 }
	 break;
	 case 11:
	 sum+=30-tt2.vec[2]+tt1.vec[2];
	 break;
	 }
	 if(tt1.vec[0]%400==0||(tt1.vec[0]%100!=0&&tt1.vec[0]%4==0))
	 {
	 if((tt2.vec[1]==1&&tt1.vec[1]!=2)||tt2.vec[1]==2)
	 sum+=1;
	 }
	 }
	}
	if(t1.compare(t2))
	 return sum;
	return -sum;
	}
	bool time::compare(const time& tt)////大于形参tt则返回true
	{
	if((vec[0]>tt.vec[0])||(vec[0]==tt.vec[0]&&vec[1]>tt.vec[1])||(vec[0]==tt.vec[0]&&vec[1]==tt.vec[1]&&vec[2]>tt.vec[2]))
	 return true;
	return false;
	}
	istream& operator>>(istream& is,time& t)
	{
	string st;
	while(getline(cin,st,'-'))
	 t.vec.push_back(stoi(st));
	cin.clear();
	return is;
	}
	ostream& operator<<(ostream& os,const time& t)
	{
	os<<t.vec[0]<<"-"<<t.vec[1]<<"-"<<t.vec[2];
	return os;
	}
	time* time::preday()
	{
	vector<int> vvv;
	if((vec[1]==5||vec[1]==7||vec[1]==10||vec[1]==12)&&(vec[2]==1))
	{
	 vvv.push_back(vec[0]);
	 vvv.push_back(vec[1]-1);
	 vvv.push_back(30);
	 return new time(vvv);
	}
	if((vec[1]==2||vec[1]==4||vec[1]==6||vec[1]==8||vec[1]==9||vec[1]==11)&&(vec[2]==1))
	{
	 vvv.push_back(vec[0]);
	 vvv.push_back(vec[1]-1);
	 vvv.push_back(31);
	 return new time(vvv);
	}
	if(vec[1]==1&&vec[2]==1)
	{
	 vvv.push_back(vec[0]-1);
	 vvv.push_back(12);
	 vvv.push_back(31);
	 return new time(vvv);
	}
	if(vec[1]==3&&vec[2]==1)
	{
	 vvv.push_back(vec[0]);
	 vvv.push_back(2);
	 if(vec[0]%400==0||(vec[0]%100!=0&&vec[0]%4==0))
	 vvv.push_back(29);
	 else
	 vvv.push_back(28);
	 return new time(vvv);
	}
	 vvv.push_back(vec[0]);
	 vvv.push_back(vec[1]);
	 vvv.push_back(vec[2]-1);
	 return new time(vvv);
	}
	bool time::test()
	{
	int da[]={1,3,5,7,8,10,12};
	int xiao[]={4,6,9,11};
	 if(find(da,da+7,vec[1])!=da+7)
	 {
	
	 if(vec[2]<1||vec[2]>31)
	 {
	 vec.clear();
	 return false;
	 }
	 return true;
	 }
	 if(vec[1]==2)
	 {
	 if(vec[0]%400==0||(vec[0]%100!=0&&vec[0]%4==0))
	 {
	 if(vec[2]<1||vec[2]>29)
	 {
	 vec.clear();
	 return false;
	 }
	 return true;
	 }
	 if(vec[2]<1||vec[2]>28)
	 {
	 vec.clear();
	 return false;
	 }
	 return true;
	 }
	 if(find(xiao,xiao+4,vec[1])!=xiao+4)
	 {
	 if(vec[2]<1||vec[2]>30)
	 {
	 vec.clear();
	 return false;
	 }
	 return true;
	 }
	 return false;
	}
	class daily
	{
	public:
	daily(const time& tt):daytime(new time(tt)),use(new int(1)){}
	daily(const daily&);
	daily& operator=(const daily&);
	void del(){if(--*use==0) delete daytime,use;}
	~daily(){del();}
	//int calculate();
	void display();
	private:
	time *daytime;
	int *use;
	static time standardtime;
	};
	time daily::standardtime(vec);
	void daily::display()
	{
	long sum=*daytime-standardtime;
	if(sum>=0)
	{
	 switch(sum%7)
	 {
	 case 0:
	 cout<<"当天为星期日"<<endl;
	 break;
	 case 1:
	 cout<<"当天为星期一"<<endl;
	 break;
	 case 2:
	 cout<<"当天为星期二"<<endl;
	 break;
	 case 3:
	 cout<<"当天为星期三"<<endl;
	 break;
	 case 4:
	 cout<<"当天为星期四"<<endl;
	 break;
	 case 5:
	 cout<<"当天为星期五"<<endl;
	 break;
	 case 6:
	 cout<<"当天为星期六"<<endl;
	 break;
	 }
	}
	else
	{
	 switch((-sum)%7)
	 {
	 case 0:
	 cout<<"当天为星期日"<<endl;
	 break;
	 case 1:
	 cout<<"当天为星期六"<<endl;
	 break;
	 case 2:
	 cout<<"当天为星期五"<<endl;
	 break;
	 case 3:
	 cout<<"当天为星期四"<<endl;
	 break;
	 case 4:
	 cout<<"当天为星期三"<<endl;
	 break;
	 case 5:
	 cout<<"当天为星期二"<<endl;
	 break;
	 case 6:
	 cout<<"当天为星期一"<<endl;
	 break;
	 }
	}
	}
	daily::daily(const daily& d)
	{
	++(*d.use);
	use=d.use;
	daytime=d.daytime;
	}
	daily& daily::operator=(const daily& d)
	{
	++(*d.use);
	del();
	daytime=d.daytime;
	use=d.use;
	return *this;
	}
	int _tmain(int argc, _TCHAR* argv[])
	{
	cout<<"日期查询界面"<<endl<<endl;
	cout<<"请输入具体日期(格式为00-00-00）"<<endl;
	time t;
	cin>>t;
	if(!t.test())
	{
	 cout<<"输入的日期有误！"<<endl;
	 return 0;
	}
	daily d(t);
	d.display();
	return 0;
	}
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