Sicily 1814. 日期计算问题

【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
 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】
#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;
}

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;
}