sicily 1814.日期计算问题

解题思路：对每个日期都计算其到0年0月0日的天数，然后两者相减.

#include <iostream>
#include <string>
using namespace std;

int longmonth[12]={31,29,31,30,31,30,31,31,30,31,30,31};//闰年每月的天数
int shortmonth[12]={31,28,31,30,31,30,31,31,30,31,30,31};//平年每月的天数

int day(int year,int month,int day)
{
int sum = 0;
for(int i = 0;i < year;i++)
{
if(i % 400 == 0 || (i % 4 == 0 && i % 100 != 0))//是闰年
sum++;
}

sum += year * 365;

if(year % 400 == 0 || (year % 4 == 0 && year % 100 != 0))//是闰年
{
for(int i = 0;i < month - 1;i++)
sum += longmonth[i];
}
else
{
for(int i = 0;i < month - 1;i++)
sum += shortmonth[i];
}

sum += day;
return sum;
}

int main()
{
int caseNum;
cin >> caseNum;
while(caseNum--)
{
string s1,s2;
cin >> s1 >> s2;

int year1 = (s1[0] - '0') * 1000 + (s1[1] - '0') * 100 + (s1[2] - '0') * 10 + (s1[3] - '0');
int month1 = (s1[5] - '0') * 10 + (s1[6] - '0');
int day1 = (s1[8] - '0') * 10 + (s1[9] - '0');

int year2 = (s2[0] - '0') * 1000 + (s2[1] - '0') * 100 + (s2[2] - '0') * 10 + (s2[3] - '0');
int month2 = (s2[5] - '0') * 10 + (s2[6] - '0');
int day2 = (s2[8] - '0') * 10 + (s2[9] - '0');

//计算每个日期与0年0月0日相差的天数
int result1 = day(year1,month1,day1);
int result2 = day(year2,month2,day2);
if(result1 - result2 > 0)
cout << result1 - result2 << endl;
else
cout << result2 - result1 << endl;
}
return 0;
}