25: calculates the number of days between two dates, and 25 calculates the number of days of a date.

Source: Internet
Author: User

25: calculates the number of days between two dates, and 25 calculates the number of days of a date.
25: calculate the number of days between two dates

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Total time limit:
1000 ms
 
Memory limit:
65536kB
Description

Calculate the number of days for the difference given two dates. For example, the difference between and is two days.

Input
Two rows in total:
The first line contains three integers: startYear, startMonth, and startDay, which are the start year, month, and day.
The second row contains three integers: endYear, endMonth, and endDay.
Two Adjacent integers are separated by a single space.

The year range is 1 ~ 3000. Ensure that the date is correct and the end date is not earlier than the start date.
Output
Returns an integer, that is, the number of days between two dates.
Sample Input
2008 1 12009 1 1
Sample output
366
Prompt
A leap year is defined as a year that can be divided by four, but can be divided by 100 instead of 400. They are not leap years. A leap year has 29 days in January.
1 # include <iostream> 2 using namespace std; 3 int bgyear, bgmonth, bgday; 4 int enyear, enmonth, enday; 5 int month [21] = {0, 31, 28, 31, 30,31, 30,31, 31,30, 31,30, 31}; // non-leap year 6 int rmonth [21] = {, 29,31, 30,31, 30,31, 31,30, 31,30, 31 }; // 7 7 int flag = 1; 8 int tot = 0; 9 int main () 10 {11 cin> bgyear> bgmonth> bgday; 12 cin> enyear> enmonth> enday; 13 for (int I = bgyear; I <= enyear + 1; I ++) // find the difference in the number of years 14 {15 if (I % 4 = 0 & I % 100! = 0) | (I % 400 = 0) 16 {17 for (int j = 1; j <= 12; j ++) 18 {19 if (I = bgyear & j <bgmonth) 20 continue; // search for the start month 21 for (int k = 1; k <= rmonth [j]; k ++) 22 {23 if (I = enyear & j = enmonth & k = enday) 24 {25 cout <tot; 26 return 0; 27} 28 if (k <bgday & flag = 1) 29 continue; 30 else31 {32 flag = 0; 33 tot ++; 34} 35 36} 37 38} 39} // leap year 40 else41 {42 43 for (int j = 1; j <= 12; j ++) 44 {45 if (I = bgyear & j <bgmonth) 46 continue; // find the start month 47 for (int k = 1; k <= month [j]; k ++) 48 {49 if (I = enyear & j = enmonth & k = enday) 50 {51 cout <tot; 52 return 0; 53} 54 if (k <bgday & flag = 1) 55 continue; 56 else57 {58 flag = 0; 59 tot ++; 60} 61 62} 63 64} 65} // non-leap year 66} 67 cout <tot; 68 return 0; 69}

 

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