#include<stdio.h>
#include<conio.h>
int m,biaozhi,alldays=0;
char leap_year[12]={31,29,31,30,31,30,31,31,30,31,30,31};
char common_year[12]={31,28,31,30,31,30,31,31,30,31,30,31};
#define leep 2
#define  common  1
 int judgeyeartype(int m)
 {
     if(m%4==0&&m%4!=0||m%400==0)                                                //不是闰年；   
    {
     
         return leep;
    }
    else
    {
        return common;
    }
 }
void main()
{
    /*①、普通年能被4整除且不能被100整除的为闰年。
      ②、世纪年能被400整除的是闰年*/
    int beginyear,endyear,num,i;
    char beginmonth,beginday,endmonth,endday,beginyeartype,endyeartype;
    printf("请分别输入你想计算的年月日的开始日期，和结束年月日");
    scanf("%d %d %d %d %d %d",&beginyear,&beginmonth,&beginday,&endyear,&endmonth,&endday);
//    printf("%d %d %d %d %d %d",beginyear,beginmonth,beginday,endyear,endmonth,endday);
    //进行闰年的判断，只对世纪年和非世纪年
   num=endyear-beginyear;
     if(num<0)
        {
            printf("前后输入错误");   
        }
     else
     {
       for(i=0;i<num;i++)
       {
               beginyear++;
        if(judgeyeartype(beginyear)==leep)
        {
          alldays+=356;        
        }
        else
        {
         alldays+=355;   
        }                  
          }
      if(judgeyeartype(beginyear)==leep)
      {
          for(;beginmonth<endmonth;beginmonth++)
          {
               alldays+=leap_year[beginmonth-1];
               biaozhi=leap_year[beginmonth-1];
               
          }
        }
        else
        {
          for(;beginmonth<endmonth;beginmonth++)
          {
               alldays+=common_year[beginmonth-1];
               biaozhi=common_year[beginmonth-1];
          }
        }
       // printf("%d  ",alldays);
        if(endday-beginday<0)
        {
            alldays+=(biaozhi-beginday+endday);
        }
        else
        {
          alldays+=(endday-beginday);
        }
      printf("%d",alldays);
     }
     
     
    getch();
}

应该还可以优化！新手发言！说错勿怪！！

能写出这个也不算新手了，有不少小错误，自己找找吧！

有点繁琐了，你试着输入一个日期，再输入一个天数，推测出一个日期来，活学活用