Sicily 1814. 日期计算问题

落日映苍穹つ 2022-08-08 00:53 51阅读 0赞

【转自网络大神，由本人整理】

Description

试用 C++的类来表示日期,给定 2 个日期 yyyy.mm.dd 求两个日期间相差的天数。

Input

第 1 行为一个正整数T，表示测试数。

对于每个测试点，第 1 行与第 2 行分别有两个日期 yyyy.mm.dd。

Output

对于每个测试点，输出一行数字，表示相差的天数。

Sample Input

![Image 1][] Copy sample input to clipboard

2
    2000.02.28
    2000.03.01
    6297.01.21
    2351.11.27

Sample Output

2
    1440938

*  计算闰年：能被4整除且不能被100整除的是闰年； 能被400整除的是闰年
 *  计算出起始日期是该年的哪一天day1
 *  计算出终止日期是该年的哪一天day2
 *  计算出两个年份之间的天数，包括起始的年份，不包括终止的年份
 *  用上一步的天数减去起始日期的所在年的day1，加上终止日期所在年的day2即为所求答案

方法一：【来自@ [chenhq1991][]】

#include<iostream>
    #include<string>
    using namespace std;
    
    int DaysOfYear(int year)
    {
    	if((year % 4 == 0 && year % 100 != 0) ||(year % 400 == 0))
    	{
    		return 366;//闰年
    	}
    	else
    	{
    		return 365;
    	}
    }
    
    int DaysOfMonth(int year,int month)
    {
    	switch(month)
    	{
    	case 1:
    	case 3:
    	case 5:
    	case 7:
    	case 8:
    	case 10:
    	case 12:
    		return 31;
    	case 2:
    		if((year % 4 == 0 && year % 100 != 0) ||(year % 400 == 0))
    			return 29;
    		else
    			return 28;
    	case 4:
    	case 6:
    	case 9:
    	case 11:
    		return 30;
    	default:
    		return 0;
    	}
    }
    class Date{
    public:
    	Date(string date)
    	{
    		year = month = day = 0;
    		int firstDot = date.find('.',0);//找到年后的那个点
    		int secondDot = date.find('.', firstDot + 1);//找到月后的那个点
    		//处理day
    		int counter = 0;
    		for(int i = date.length() - 1; i > secondDot; i--)
    		{
    			char d = date[i];
    			int temp = d - 48;
    			for(int j = 0; j < counter; j++)
    			{
    				temp = temp * 10;
    			}
    			day = day + temp;
    			counter++;	
    		}
    		counter = 0;
    		//处理month
    		for(int i = secondDot - 1; i > firstDot; i--)
    		{
    			char m = date[i];
    			int temp = m - 48;
    			for(int j = 0; j < counter; j++)
    			{
    				temp = temp * 10;
    			}
    			month = month + temp;
    			counter++;
    		}
    		counter = 0;
    		//处理year
    		for(int i = firstDot - 1; i >= 0; i--)
    		{
    			char y = date[i];
    			int temp = y - 48;
    			for(int j = 0; j < counter; j++)
    			{
    				temp = temp * 10;
    			}
    			year = year + temp;
    			counter++;
    		}
    	}
    	int GetYear()
    	{
    		return year;
    	}
    
    	int GetMonth()
    	{
    		return month;
    	}
    
    	int GetDay()
    	{
    		return day;
    	}
    	//重载‘-’
    	//要求left 的时间晚于 right的
    	friend int operator - (Date left, Date right)
    	{
    		int daysBeforeRight = 0;
    		//先计算出右边的日期是一年中的哪一天
    		for(int i = 1; i < right.GetMonth(); i++)
    		{
    			daysBeforeRight = daysBeforeRight + DaysOfMonth(right.GetYear(), i);
    		}
    		daysBeforeRight = daysBeforeRight + right.GetDay();//如果是2000.1.1，daysBeforeRight 就是1
    		int daysBeforeLeft = 0;
    		//计算出左边的日期是一年中的哪一天
    		for(int i = 1; i < left.GetMonth(); i++)
    		{
    			daysBeforeLeft = daysBeforeLeft + DaysOfMonth(left.GetYear(), i);
    		}
    		daysBeforeLeft = daysBeforeLeft + left.GetDay();
    		int interval = 0;//两个日期的间隔
    		for(int i = right.GetYear(); i < left.GetYear(); i++)
    		{
    			interval = interval + DaysOfYear(i);
    		}
    		//实际的间隔
    		interval = interval - daysBeforeRight + daysBeforeLeft;
    		return interval;
    	}
    private:
    	int year;
    	int month;
    	int day;
    };
    
    int main()
    {
    	int caseNum;
    	cin >> caseNum;
    	while(caseNum--)
    	{
    		string first, second;
    		cin >> first >> second;
    		Date one(first);
    		Date two(second);
    		//比较两个时间的先后
    		if(first.length() > second.length())//长度长的时间较晚
    		{
    			int interval = one - two;
    			cout << interval << endl;
    		}
    		else
    		{
    			if(first.length() < second.length())
    			{
    				int interval = two - one;
    				cout << interval << endl;
    			}
    			else//长度相等
    			{
    				if(first > second)//字符串比较大小，大的时间较晚
    				{
    					int interval = one - two;
    					cout << interval << endl;
    				}
    				else
    				{
    					int interval = two - one;
    					cout << interval << endl;
    				}
    			}
    		}
    		
    	}
    	return 0;
    }

方法二：【来自@ [绿夜][Link 1]】

自己的，基本按照大神的思路

#include<iostream>
    #include<cstdio>
    #include<cctype>
    #include<iomanip>
    #include<vector>
    #include<cstring>
    #include<string>
    #include<fstream>
    #include<stack>
    #include<vector>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    
     bool runnian(int year){
        if(year%400==0 || (year%4==0 && year%100 != 0)){
            return true;
        }
        else return false;
    }
     int year1,mouth1,day1,year2,mouth2,day2;
    
     int longmouth[12]={31,29,31,30,31,30,31,31,30,31,30,31};
    
     int shortmouth[12]={31,28,31,30,31,30,31,31,30,31,30,31};
    
     long days(int year,int mouth,int day){
        long amounts = 0;
        int i,j;
        for(i=0;i<year;i++){
            if(runnian(i))amounts++;
        }
        amounts+=year*365;
        if(runnian(year)){
            for(i=0;i<mouth-1;i++)
                amounts+=longmouth[i];
        }
        else{ 
            for(i=0;i<mouth-1;i++)
                amounts+=shortmouth[i];
        }
        amounts+=day;
        return amounts;
    }
    
     int main(){
        int t;
        long d;
        char str[11];
        cin>>t;
        while(t--){
            cin>>str;
            year1=(str[0]-'0')*1000+(str[1]-'0')*100+(str[2]-'0')*10+(str[3]-'0');
            mouth1=(str[5]-'0')*10+(str[6]-'0');
            day1=(str[8]-'0')*10+(str[9]-'0');
            cin>>str;
            year2=(str[0]-'0')*1000+(str[1]-'0')*100+(str[2]-'0')*10+(str[3]-'0');
            mouth2=(str[5]-'0')*10+(str[6]-'0');
            day2=(str[8]-'0')*10+(str[9]-'0');
            if((d=days(year1,mouth1,day1)-days(year2,mouth2,day2))>0)
                cout<<d<<endl;
            else cout<<-d<<endl;
        }
        return 1;
    }

[Image 1]: 
[chenhq1991]: http://my.csdn.net/chenhq1991
[Link 1]: http://home.cnblogs.com/u/lvye/