根据日期求星期（1900年及以后）

#include <stdio.h>
int main()
{
 int a,b,c,year,day,leap,d;
 scanf("%d%d%d",&a,&b,&c);
 year=a-1900;
 leap=year/4;
 switch(b)
 {
    case 1:
    if(0==(year%4)) day=year*365+leap+c-1;
     else day=year*365+leap+c;
    break;
    case 2:
    if(0==(year%4)) day=year*365+leap+30+c;
    else day=year*365+leap+31+c;
    break;
    case 3:
    day=year*365+leap+59+c;
    break;
    case 4:case 5:
    day=year*365+leap+30*(b-1)+c;
    break;
    case 6:case 7:
    day=year*365+leap+30*(b-1)+1+c;
    break;
    case 8:
    day=year*365+leap+30*(b-1)+2+c;
    break;
    case 9:case 10:
    day=year*365+leap+30*(b-1)+3+c;
    break;
    case 12:
    if (a!=1900) day=year*365+leap+30*(b-1)+4+c;
     else day=334+c;
    break;
    case 11:
    day=year*365+leap+30*(b-1)+4+c;
    break;
 }
 d=day%7;
 switch(d)
{
 case (0):
 d=0;
 printf("%d\n",d);
 break;
 default:
 printf("%d\n",d);
 break;
}

}