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Week 16 OJ practice 1 calculates the day of the year

2022-07-26 17:31:00 【51CTO】

Problems and code ：

Problem A: C Language exercises Calculate the day of the year

       
       Time Limit: 1 Sec  
       
 Memory Limit: 128 MB
       
Submit: 3065  
       
 Solved: 1670
       
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Description

Define a structure variable （ Including the year 、 month 、 Japan ）. Write a function days, By the main function, the year 、 month 、 Day passed to function days, Calculate the day of the year and send the result back to the main function output .

Input

Specific date

Output

The first few days of that year

Sample Input

Sample Output

       
       /*
       
        School of computer science, Yantai University 
       
        author ： Jingyile 
       
        Completion time ：2016 year 12 month 16 Japan 
       
       */
       
       #include <stdio.h>
       
       struct 
       
       y_m_d
       
{
       
       int 
       
       year;
       
       int 
       
       month;
       
       int 
       
       day;
       
}
       
       date;
       
       // Define structure variables 
       
       int 
       
       main()
       
{
       
       struct 
       
       y_m_d 
       
       date;
       
       int 
       
       days(
       
       y_m_d);
       
       int 
       
       day_sum;
       
       scanf(
       
       "%d%d%d",
       
       &
       
       date.
       
       year,
       
       &
       
       date.
       
       month,
       
       &
       
       date.
       
       day);
       
       day_sum
       
       =
       
       days(
       
       date);
       
       printf(
       
       "%d\n",
       
       day_sum);
       
       return 
       
       0;
       
}
       
       int 
       
       days(
       
       struct 
       
       y_m_d 
       
       date)
       
{
       
       int 
       
       sum
       
       =
       
       date.
       
       day;
       
       int 
       
       a[
       
       13]
       
       = {
       
       0,
       
       31,
       
       28,
       
       31,
       
       30,
       
       31,
       
       30,
       
       31,
       
       31,
       
       30,
       
       31,
       
       30,
       
       31};
       
       int 
       
       i;
       
       for(
       
       i
       
       =
       
       1; 
       
       i
       
       <
       
       date.
       
       month; 
       
       i
       
       ++)
       
    {
       
       sum
       
       +=
       
       a[
       
       i];
       
    }
       
       if(
       
       date.
       
       month
       
       >
       
       2
       
       &&((
       
       date.
       
       year
       
       %
       
       4
       
       ==
       
       0
       
       &&
       
       date.
       
       year
       
       %
       
       100
       
       !=
       
       0)
       
       ||
       
       date.
       
       year
       
       %
       
       400
       
       ==
       
       0)) 
       
       // If leap year , And later 2 month , Add a day 
       
       sum
       
       ++;
       
       return 
       
       sum;
       
}
      
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Running results ：

The first 16 Zhou OJ practice 1 Calculate the day of the year _ to be divisible by

Summary of knowledge points ： You can define an array , First press February 28 Days counted , Finally, judge whether it is a leap month , if , Add a day , Remember to also judge that it is after February .

How to judge leap years
The rule that the leap year of the Gregorian calendar follows is :

A leap in four years , A hundred years without Leap , Four hundred years later .
A simple calculation method of leap year in Gregorian calendar （ A leap year is one of the following ）

1. Can be 4 Divide without being 100 to be divisible by .

2. Can be 400 to be divisible by .

The learning ： Once again, I encountered the problem about leap years , Leap year is judged by (n%4==0&&n%100!=0)||(n%400==0)

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