年月日问题

方法一：利用"swith"语句

//输入一个日期，如“2013 8 8”,输出该日期是当年第几天。
#include"stdio.h"
void main()
{
    int year,month,day,sum,leap;
    sum=0;
    printf("input the date:");
    scanf("%d%d%d",&year,&month,&day);
    switch(month)
    {
        case 12:sum+=30; //注意：自加的为上个月的天数；
        case 11:sum+=31;
        case 10:sum+=30;
        case 9:sum+=31;
        case 8:sum+=31;
        case 7:sum+=30;
        case 6:sum+=31;
        case 5:sum+=30;
        case 4:sum+=31;
        case 3:sum+=28;
        case 2:sum+=31;
        case 1:sum+=day;break;
        default:printf("Error!\n");
    }
    if((year%4==0&&year%100!=0)||year%400==0) //判断闰年
        leap=1;
    else
        leap=0;;
    if(leap&&month>2) //注意：month>2
        sum++;
    printf("The day is the %dth day of the year!\n",sum);
}

 

方法二：构造函数

//输入一个日期，如“2013 1 4”,输出该日期是当年第几天。
#include"stdio.h"
int main()
{
    int day_sum(int month,int day);
    int leap(int year);
    int year,month,day,days;
    printf("Please input data:");
    scanf("%d%d%d",&year,&month,&day);
    days=day_sum(month,day);
    if(leap(year)==1&&month>2)
        days=days+1;
    printf("%d/%d/%d is the %dth day in this year.\n",year,month,day,days);
    return 0;
}



//计算天数函数
int day_sum(int month,int day)
{
    int day_tab[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
    int i;
    for(i=1;i<month;i++)
        day=day+day_tab[i];
    return day;
}
//判断闰年函数
int leap(int year)
{
    int leap;
    leap=year%4==0&&year%100!=0||year%400==0;
    return leap;
}

方法三：利用结构体变量

//定义一个结构体变量（包括年、月、日）。计算该日在本年中是第几天，注意闰年问题。
#include"stdio.h"
struct
{
    int year;
    int month;
    int day;
}Date;
void main()
{
    int data[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
    int i,n,day,days=0;
    printf("Please input the date:");
    scanf("%d%d%d",&Date.year,&Date.month,&Date.day);
    n=Date.month;
    for(i=1;i<n;i++)
        days+=data[i];
    if(((Date.year%4==0&&Date.year%100!=0)||Date.year%400==0)&&Date.month>2)
        day=days+Date.day+1;
    else
        day=days+Date.day;
    printf("The day is the %dth day of the year.\n",day);
}