hdu 2005 asking for the day

Topic link: http://acm.hdu.edu.cn/showproblem.php?pid=2005

                                        What day?

                          Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                   Total Submission(s): 177873    Accepted Submission(s): 63023

Problem Description

Given a date, output the day of the year the date is.

 

Input

There are multiple groups of input data, each group occupies one line, and the data format is YYYY/MM/DD. For details, please refer to sample input. In addition, you can ensure that all input data are legal.

 

Output

For each set of input data, output a row indicating the day of the year the date is.

 

Sample Input

1985/1/20

2006/3/12

 

Sample Output

20

71

 

#include <stdio.h>  

int leapyear_day(int year, int month)
{
    // do not need to add 1 day to January or February, add 1 day to other months, and add 1 day to non-money   
    if (month <= 2 )
         return  0 ;
     else 
        return (((year % 4 == 0 ) && (year % 100 != 0 )) || (year % 400 == 0 )) ? 1 : 0 ;
}

int main(void)
{
    int year, month, day;
     int days;
     int monthsum[] = {   // The number of days to a certain month, plus the number of days to run the year   
        0                                  // January   
        , 31                                 // February   
        , 31 + 28                              // March   
        , 31 + 28 + 31                           // April   
        , 31 + 28 + 31 + 30                        // May   
        , 31 + 28 + 31 + 30 +31                     // June   
        , 31 + 28 + 31 + 30 + 31 + 30                  // July   
        , 31 + 28 + 31 + 30 + 31 + 30 + 31               // August   
        , 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31            // September   
        ,31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30         // October   
        , 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31      // November   
        , 31 + 28 + 31 + 30 + 31 + 30 +31 + 31 + 30 + 31 + 30   // December   
    };

    while (scanf( " %d/%d/%d " , &year, &month, &day) != EOF) {
         // The number of days = the number of days to be added to the year (calculated according to the year and month) + the number of days to a certain month + the month Inner days   
        days = leapyear_day(year, month) + monthsum[month - 1 ] + day;

        // output result   
        printf( " %d\n " , days);
    }

    return 0;
}

 

 

2018-04-23