main()
{long a,a1,a2;
long leap=0,b,c,d,e,f,i=0,j;
long dd[10000];
scanf("%ld",&dd[0]);
while(dd[i]!=-1)
{i++;
scanf("%ld",&dd[i]);
}
for(i=0;dd[i]!=-1;i++)
{dd[i]+=1;
a=dd[i]/365;
b=dd[i]%365;
c=a/4;
d=a/100;
e=a/400;
f=c-d+e;
j=0;
while(f>b)
{j++;
b+=365;
}
a-=j;
b-=f;
c=a%4;
d=a%100;
e=a%400;
if((!c&&d)||!e)leap=1;
if(b<=31){a1=1;a2=b;}
else if(b<=59+leap){a1=2;a2=b-31;}
else if(b<=90+leap){a1=3;a2=b-59-leap;}
else if(b<=120+leap){a1=4;a2=b-90-leap;}
else if(b<=151+leap){a1=5;a2=b-120-leap;}
else if(b<=181+leap){a1=6;a2=b-151-leap;}
else if(b<=212+leap){a1=7;a2=b-181-leap;}
else if(b<=243+leap){a1=8;a2=b-212-leap;}
else if(b<=273+leap){a1=9;a2=b-243-leap;}
else if(b<=304+leap){a1=10;a2=b-273-leap;}
else if(b<=334+leap){a1=11;a2=b-304-leap;}
else {a1=12;a2=b-334-leap;}
a+=2000;
printf("%ld-",a);
if(a1<10)printf("0%ld-",a1);
else printf("%ld-",a1);
if(a2<10)printf("0%ld ",a2);
else printf("%ld ",a2);
switch(dd[i]%7)
{case 1 :printf("Saturday\n");break;
case 2 :printf("Sunday\n");break;
case 3 :printf("Monday\n");break;
case 4 :printf("Tuesday\n");break;
case 5 :printf("Wednesday\n");break;
case 6 :printf("Thursday\n");break;
case 0 :printf("Friday\n");break;
}
}
}

• 利用julian date可以有一套连一个if都没有的公式，去找一下吧 (1355) frkstyc 2005-05-27 16:31:58
  • Re:利用julian date可以有一套连一个if都没有的公式，去找一下吧 (202) wdz 2006-07-26 13:52:31
    • Re:利用julian date可以有一套连一个if都没有的公式，去找一下吧 (14) xinjin99 2007-07-22 19:54:51
• 用JAVA吧*^_^* (1355) FinalLaugh 2005-05-27 19:10:39
• 当 输入 0 时候是 周几呀？？？？？？？？？？？ (1355) tzj115 2007-07-28 23:33:05