[YTU]_2428(C语言习题 计算该日在本年中是第几天)

题目描述

定义一个结构体变量（包括年、月、日）。编写一个函数days，由主函数将年、月、日传递给函数days，计算出该日在本年中是第几天并将结果传回主函数输出。

输入

年月日

输出

当年第几天

样例输入

2000 12 31

样例输出

366
#include <iostream>
using namespace std;
struct y_m_d
{
    int year,month,day;
};
int days(y_m_d date)
{
    int m;
    if((date.year%4==0&&date.year%100!=0)||(date.year%400==0))
        switch(date.month)
    {
        case 1:m=date.day;break;
        case 2:m=31+date.day;break;
        case 3:m=31+29+date.day;break;
        case 4:m=31+29+31+date.day;break;
        case 5:m=31+29+31+30+date.day;break;
        case 6:m=31+29+31+30+31+date.day;break;
        case 7:m=31+29+31+30+31+30+date.day;break;
        case 8:m=31+29+31+30+31+30+31+date.day;break;
        case 9:m=31+29+31+30+31+30+31+31+date.day;break;
        case 10:m=31+29+31+30+31+30+31+31+30+date.day;break;
        case 11:m=31+29+31+30+31+30+31+31+30+31+date.day;break;
        case 12:m=31+29+31+30+31+30+31+31+30+31+30+date.day;break;
    }
    else
        switch(date.month)
    {
        case 1:m=date.day;break;
        case 2:m=31+date.day;break;
        case 3:m=31+28+date.day;break;
        case 4:m=31+28+31+date.day;break;
        case 5:m=31+28+31+30+date.day;break;
        case 6:m=31+28+31+30+31+date.day;break;
        case 7:m=31+28+31+30+31+30+date.day;break;
        case 8:m=31+28+31+30+31+30+31+date.day;break;
        case 9:m=31+28+31+30+31+30+31+31+date.day;break;
        case 10:m=31+28+31+30+31+30+31+31+30+date.day;break;
        case 11:m=31+28+31+30+31+30+31+31+30+31+date.day;break;
        case 12:m=31+28+31+30+31+30+31+31+30+31+30+date.day;break;
    }
    return m;
}
 
int main()
{
    y_m_d date;
    int days(y_m_d);  
    int day_sum;
    cin>>date.year>>date.month>>date.day;
    day_sum=days(date);
    cout<<day_sum<<endl;
    return 0;
}