输入某年某月的某日，判断这天是改年的第多少天，
{要求用函数实现？？？}



急求!!!!!1111

为什么不自己写呢

会就不问了；
会就不请教了；

这个程序是我自己写的，比较繁琐，但算法还是比较简单的，楼主可以在我的基础上加以改进。

#include "iostream.h"
#include "stdio.h"

#define COUNT  printf("It is error!\n")

int NoLeap_day(int year);
int Leap_day(int year);

struct date
{   
    int year;
    int months[2][13];
    int day;
}DATE;

void main()
{
    struct date DATE;
    struct date *p;
    p=&DATE;
    int Dayst;
    cin>>p->year;
    if((p->year%100!=0)&&(p->year%4==0)||(p->year%400==0))
    {
        Dayst=Leap_day(p->year);
    }
    else
    {
        Dayst=NoLeap_day(p->year);
    }
}

int Leap_day(int year)
{
    int i,month,day,Dayst,months[2][13]={{0,31,29,31,30,31,30,31,31,30,31,30,31},{0,31,28,31,30,31,30,31,31,30,31,30,31}};
    cin>>month>>day;
    Dayst=0;
    if(month<=12)
    {
    if(day>months[0][month])
    {
        return COUNT;
    }
    else
    {
    for(i=0;i<month;i++)
    {
        Dayst+=months[0][i];

    }
    Dayst+=day;
    return     printf("The day is the %d day.",Dayst);
    }
    }
    else return COUNT;
}

int NoLeap_day(int year)
{
int i,month,day,Dayst,months[2][13]={{0,31,29,31,30,31,30,31,31,30,31,30,31},{0,31,28,31,30,31,30,31,31,30,31,30,31}};
    cin>>month>>day;
    Dayst=0;
    if(month<=12)
    {
    if(day>months[0][month])
    {
        return COUNT;
    }
    else
    {
    for(i=0;i<month;i++)
    {
        Dayst+=months[1][i];
    }
    Dayst+=day;
    return     printf("The day is the %d day.",Dayst);
    }
    }
    else return COUNT;
}