The idea of calculating the number of days between two dates:

First, determine whether the year entered is a leap years. Is the year the same? Is the month the same? Is the day the same?

There are three possible scenarios for the sun and moon year:

- Same year in the same month. The number of days to subtract is out.
- Same year different months. Calculates the number of days in a small month to the beginning of the day, and calculates the number of days from the month to the beginning. Two more dates to reduce
- Different years. Calculate the middle of a few years apart, calculate the number of days to the end of a smaller date, and then calculate the number of days from the beginning of the larger date, and add three numbers to it.

The code is as follows:

1#include <iostream>2#include <CString>3#include <cmath>4 using namespacestd;5 6 BOOLIsleapyear (intYear//determine if a leap year7 {8 if(Year%4==0&& Year% -!=0|| Year% -==0)9 return 1;Ten Else One return 0; A } - - intDays (intYintMintD//Calculate y years m months D days to Y year January 1 days the { - intDays[] = {0, to, -, to, -, to, -, to, to, -, to, -, to }; - inti; - intsum =0;/*Count Days*/ + if(Isleapyear (y))/*if it is a leap year, February has 29 days*/ -days[2] = in; + for(i =0; i<m; i++) Asum = sum +Days[i]; atsum = sum + D-1; - returnsum; - } - - intDays (intY1,intM1,intD1,intY2,intM2,intD2)//calculates the number of days between two dates that are not the same as the month and day - { in intT1; - intT2; to intYear = y2-Y1; + intsum=0;//Days of difference - if(Year = =1)//if the year of two dates differs by 1 the { *T1 = Days (y1, A, to)-Days (Y1, M1, D1);//Smaller date calculates Y1 year M1 month D1 day to end of days $T2 = days (y2, M2, D2);//A Larger date calculates the number of days y2 m2 months D2 days to the beginning of the yearPanax Notoginsengsum = t1 + t2 +Year ; - } the Else{ + for(inti = y1+1; i < Y2; i++) A { the if(Isleapyear (i)) + { -Sum + =366; $ } $ Else -Sum + =365; - } theT1 = Days (y1, A, to) -Days (Y1, M1, D1); -t2 =Days (y2, M2, D2);Wuyisum = sum + T1 + t2 +1; the - } Wu returnsum; - } About $ intMain () { - intyear1;//smaller year - intyear2; - intMonth1;//the smaller month A intmonth2; + intDay1; the intDay2; - intday=0; $CIN >> year1 >> month1 >>Day1; theCIN >> year2 >> month2 >>Day2; the the if(Year1 > Year2 | | (year1 = = year2) && (month1 > month2) | | (year1 = = year2) && (month1 = = month2)) && (Day1 >day2)) the { -cout <<"Input Error"<<Endl; in } the Else{ the About if(Year1 = = Year2&&month1 = = month2)//1. If the year and month are the same, calculate the difference of the number of days directly the { theDay = ABS (Day2-day1); thecout << Day <<Endl; + } - Else if(year1 = = Year2 && month1! = month2)//2. If the year is the same and the month is different, calculate the number of days from the smaller date to the larger date the {BayiDay = ABS (Days (YEAR1, Month1, year1)-Days (Year2, Month2, Day2)); thecout << Day <<Endl; the } - Else - { theDay =Days (year1, Month1, Day1, Year2, Month2, day2); thecout << Day <<Endl; the } the } - return 0; the}

For the understanding of some code:

intDays (intYintMintD//Calculate y years m months D days to Y year January 1 days{ intDays[] = {0, to, -, to, -, to, -, to, to, -, to, -, to }; inti; intsum =0;/*Count Days*/ if(Isleapyear (y))/*if it is a leap year, February has 29 days*/days[2] = in; for(i =0; i<m; i++) Sum= Sum +Days[i]; Sum= sum + D-1; returnsum;}

The understanding of this code can be expressed by:

We need to ask for a date of two red arrows between the green section.

1 for(inti = y1+1; i < Y2; i++)2 {3 if(Isleapyear (i))4 {5Sum + =366;6 }7 Else8Sum + =365;9 }TenT1 = Days (y1, A, to) -Days (Y1, M1, D1); Onet2 =Days (y2, M2, D2); Asum = sum + T1 + t2 +1;

The understanding of this code can be expressed by:

2015 with 2011 years of difference in fact only three years, just start writing when the number of years directly written is 2015 minus 2011, resulting in more than a year.

C + + calculates the number of days between two dates