nyoj 75 日期问题

#include<stdio.h>
int main()
{
int n,year,month[13],day,inputmonth,total,i;month[0] = 0;
month[1] = 31,month[2] = 28,month[3] = 31,month[4] = 30,month[5] = 31,month[6] = 30;
month[7] = 31,month[8] = 31,month[9] = 30,month[10] = 31,month[11] = 30,month[12] = 31;
scanf("%d",&n);
while(n--)
{
total = 0;
scanf("%d%d%d",&year,&inputmonth,&day);
for(i = 1;i < inputmonth;i++)
{
total += month[i] ;

}
total += day;
if(year%400 == 0 && inputmonth > 1)
{
printf("%d\n",++total);
}
else if(year%4 == 0 && year%100 != 0 && inputmonth > 1)
{
printf("%d\n",++total);
}
else
{
printf("%d\n",total);
}

}
return 0;

}

这个题我忘了考虑月份是1的情况，前几次一直没有AC过。下面是最优代码：

   
#include<stdio.h>
int main()
{
int a,b=0,c,y,m,d,fib;
scanf("%d",&a);
while(a--)
{
scanf("%d %d %d",&y,&m,&d);
if(y%400==0||y%100!=0&&y%4==0)
fib=29;
else fib=28;
for(c=1;c<=m;c++)
switch(c-1)
{
           case 1:
  case 3:
  case 5:
  case 7:
  case 8:
  case 10:b+=31;break;
  case 2:b+=fib;break;
  case 4:
  case 6:
  case 9:
  case 11:b+=30;break;
}
b+=d;
printf("%d\n",b);
b=0;
}
return 0;
}