Gardon的18岁生日就要到了，他当然很开心，可是他突然想到一个问题，是不是每个人从出生开始，到达18岁生日时所经过的天数都是一样的呢？似乎并不全都是这样，所以他想请你帮忙计算一下他和他的几个朋友从出生到达18岁生日所经过的总天数，让他好来比较一下。

 

Input
一个数T，后面T行每行有一个日期，格式是YYYY-MM-DD。如我的生日是1988-03-07。

 

Output
T行，每行一个数，表示此人从出生到18岁生日所经过的天数。如果这个人没有18岁生日，就输出-1。

 

Sample Input
1
1988-03-07
 

Sample Output
6574
 











#include<stdio.h>
int main ()
{
    int year,month,day,i,j,sum,n;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d-%d-%d",&year,&month,&day);
        
        if(year+18<2010)
        {
            if((year%400==0)||(year%4==0&&year%100!=0))
            {
                if(month==2&&day==29)
                {
                    printf("-1");
                }
            }
            else
            {
                for(i=year;i<=year+18;i++)
                {
                    if((i%400==0)||(i%4==0&&i%100!=0))
                        j++;
                }
                sum=365*18+j;
                printf("%d",sum);            
            }
            
        }
        else
        {
            printf("-1");
        }   
        
        
    }
   
   
    return 0;
}

#include<stdio.h>
int main ()
{
    int year,month,day,i,j,sum,n;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d-%d-%d",&year,&month,&day);
        
        if(year+18<2010)
        {
            if((year%400==0)||(year%4==0&&year%100!=0))
            {
                if(month==2&&day==29)
                {
                    printf("-1");
                }
            }
            
            for(i=year,j=0;i<=year+18;i++)
            {
                if((i%400==0)||(i%4==0&&i%100!=0))
                    j++;
            }
            sum=365*18+j-1;
            printf("%d\n",sum);            
            
            
        }
        else
        {
            printf("-1");
        }   
        
        
    }
   
   
    return 0;
}

当然是错了....

？？？？？题目程序都在这了
不太懂你说什么

#include<stdio.h>
int main ()
{
    int year,month,day,i,j,sum,n;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d-%d-%d",&year,&month,&day);
        
        if(year+18<=2010)
        {
            if((year%400==0)||(year%4==0&&year%100!=0))
            {
                if(month==2&&day==29)
               
                    printf("-1");
               
            }
            
            for(i=year,j=0;i<=year+18;i++)
            {
                if((i%400==0)||(i%4==0&&i%100!=0)&&month>2)
                    j++;
            }
            sum=365*18+j;
            printf("%d\n",sum);            
            
            
        }
        else
        {
            printf("-1");
        }   
        
        
    }
   
   
    return 0;
}