日期相关的题目

1. 计算日期到天数转换

输出示例:

思路：计算前n-1个月的天数在加上这个月的天数。

#include <iostream>
using namespace std;

int main() {
	int year, month, day;
	cin >> year >> month >> day;
	int month_day[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
	int i = 0;
	int days = 0;
	for (i = 0; i < month; i++)
	{
		days += month_day[i];
	}

	days += day;
    if(month > 2 && (((year % 4 == 0) && (year % 100 != 0)) || (year % 400) == 0))
	{
		days += 1;
	}

	cout << days;

	return 0;
}

2. 日期累加

输出示例:

思路：直接加到天上，在进位求出日期

#include <iostream>
using namespace std;

int GetMonthDay(int year ,int month)
{
     int months[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };


    int day = months[month];

    if (month == 2 &&( ((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0)))
    {
        day++;
    }

    return day;
}

int main() {
    int n = 0;
    cin >> n;

    int year = 0,month = 0, day = 0,num;

    while (n-- && (cin >> year >> month >> day >> num) ) 
    { 
        day += num;
        while (day > GetMonthDay(year,month))
        {
            day -= GetMonthDay(year,month);
            month++;
            if (month == 13)
            {
                month = 1;
                year++;
            }
        }

       printf("%04d-%02d-%02d\n", year, month, day);
    }
}

3. 打印日期

输出示例:

思路：从这年的1月1开始加，但是注意要要减掉一天，比如：求这一年的第一天的日期，就是1月1号，也可以理解为从1月0号开始加，所以直接加到天上，在进位求日期

#include <iostream>
using namespace std;

int GetMonthDay(int year,int month)
{
    int months[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

    int day = months[month];

   if (month == 2 && ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0) ))
    {
        day += 1;
    }

    return day;
}

int main() {
    int year = 0;
    int n = 0;
    while (cin >> year >> n) { 
        //从这年的1月1开始加,但是要减一天
        int month = 1;
        int day = n;//1 + n - 1
       while (day > GetMonthDay(year,month) )
       {
        
          day -= GetMonthDay(year,month);
          month++;
          if (month == 13)
          {
            year++;
            month = 1;
          }
       }

       printf("%4d-%02d-%02d\n",year,month,day);

    }
}

4. 日期差值

输出示例:

思路：直接从这天开始加，加到相等为止，每加一天就记一次数。

#include <iostream>
using namespace std;

int GetMonthDay(int year,int month)
{
    int months[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

    int day = months[month];

    if (month == 2 && (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0)))
    {
        day++;
    }

    return day;
}

void Add(int& year,int& month,int& day)
{
    day += 1;
    while (day > GetMonthDay(year,month))
    {
        day -= GetMonthDay(year,month);
        month++;
        if (month == 13)
        {
            month = 1;
            year++;
        }
    }
} 

int main() {
    int year1,month1,day1;
    scanf("%4d%2d%2d",&year1,&month1,&day1);

    int year2,month2,day2;
    scanf("%4d%2d%2d",&year2,&month2,&day2);

    int count = 0;
    int flag = 0;

    if (year2 > year1)
    {
        flag = 1;
    }
    else if (year1 == year2 && month1 < month2)
    {
         flag = 1;
    }
    else if (year1 == year2 && month1 == month2 && day1 < day2)
    {
        flag = 1;
    }

    if (flag == 1)
    {
        swap(year1,year2);
        swap(month1,month2);
        swap(day1,day2);
    }

    while (!(year1 == year2 && month1 == month2 && day1 == day2))
    {
        Add(year2,month2,day2);
        count++;
    }

    cout << count + 1;
}