输入年，月，日，；输出星期几
#include <stdio.h>
main()
{
    int y,m,d,w;
    scanf("%d%d%d",&y,&m,&d);
    if((y%4==0&&y%100!=0)||y%400==0)
    {
        switch(m)
        {
        case 1:w=(365*y+(y/4-y/100+y/400)+d-2)%7;break;
case 2:w=(365*y+(y/4-y/100+y/400)+d+29)%7;break;
case 3:w=(365*y+(y/4-y/100+y/400)+d+58)%7;break;
case 4:w=(365*y+(y/4-y/100+y/400)+d+89)%7;break;
case 5:w=(365*y+(y/4-y/100+y/400)+d+119)%7;break;
case 6:w=(365*y+(y/4-y/100+y/400)+d+150)%7;break;
case 7:w=(365*y+(y/4-y/100+y/400)+d+180)%7;break;
case 8:w=(365*y+(y/4-y/100+y/400)+d+211)%7;break;
case 9:w=(365*y+(y/4-y/100+y/400)+d+242)%7;break;
case 10:w=(365*y+(y/4-y/100+y/400)+d+272)%7;break;
case 11:w=(365*y+(y/4-y/100+y/400)+d+303)%7;break;
case 12:w=(365*y+(y/4-y/100+y/400)+d+333)%7;
        }
    }
        else
        {
        switch(m)
    {

    case 1:w=(365*y+(y/4-y/100+y/400)+d-1)%7;break;
    case 2:w=(365*y+(y/4-y/100+y/400)+d+30)%7;break;
    case 3:w=(365*y+(y/4-y/100+y/400)+d+58)%7;break;
    case 4:w=(365*y+(y/4-y/100+y/400)+d+89)%7;break;
    case 5:w=(365*y+(y/4-y/100+y/400)+d+119)%7;break;
    case 6:w=(365*y+(y/4-y/100+y/400)+d+150)%7;break;
    case 7:w=(365*y+(y/4-y/100+y/400)+d+180)%7;break;
    case 8:w=(365*y+(y/4-y/100+y/400)+d+211)%7;break;
    case 9:w=(365*y+(y/4-y/100+y/400)+d+242)%7;break;
    case 10:w=(365*y+(y/4-y/100+y/400)+d+272)%7;break;
    case 11:w=(365*y+(y/4-y/100+y/400)+d+303)%7;break;
    case 12:w=(365*y+(y/4-y/100+y/400)+d+333)%7;
    }
        }
    switch(w)
    {
    case 1:printf("1\n");break;
case 2:printf("2\n");break;
case 3:printf("3\n");break;
case 4:printf("4\n");break;
case 5:printf("5\n");break;
case 6:printf("6\n");break;
case 0:printf("7\n");
    }
}

检测过了好像没出错，是不是LZ输入的时候少按了回车了啊

运行起来貌似没有问题。但是细节性还是要注意的。

似乎没有什么问题，就是代码太复杂了．
应该可以简化一下．

建议：首先将判断是否是闰年的功能独立成一个函数；
其次定义两个枚举类来分别来记录月份和星期；
最后将计算是星期几的功能也独立成一个模块．

这样层次比较清晰．

没问题，把那个 scanf("%d,%d,%d",&y,&m,&d);改成这样。再输入，能够得到结果。

用c++6.0运行了下 都没出错啊

基姆拉尔森计算公式
W= (d+2*m+3*(m+1)/5+y+y/4-y/100+y/400) mod 7
  在公式中d表示日期中的日数，m表示月份数，y表示年数。除法取整。
注意：在公式中有个与其他公式不同的地方：
 把一月和二月看成是上一年的十三月和十四月，例：如果是2004-1-10则换算成：2003-13-10来代入公式计算。

那你们也帮我解释一下w=(365*y(y/4-y/100+y/400)+d-1)%7是什么意思,为啥要这样写.

#include <stdio.h>
main()
{
    int y,m,d,w;
    scanf("%d%d%d",&y,&m,&d);
    if((y%4==0&&y%100!=0)||y%400==0)
        switch(m)
        {
        case 1:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d)%7;break;
case 2:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+31)%7;break;
case 3:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+60)%7;break;
case 4:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+91)%7;break;
case 5:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+121)%7;break;
case 6:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+152)%7;break;
case 7:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+182)%7;break;
case 8:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+213)%7;break;
case 9:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+244)%7;break;
case 10:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+274)%7;break;
case 11:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+305)%7;break;
case 12:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+335)%7;
        }
        else
        switch(m)
    {

    case 1:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d)%7;break;
    case 2:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+31)%7;break;
    case 3:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+59)%7;break;
    case 4:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+90)%7;break;
    case 5:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+120)%7;break;
    case 6:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+151)%7;break;
    case 7:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+181)%7;break;
    case 8:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+212)%7;break;
    case 9:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+243)%7;break;
    case 10:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+273)%7;break;
    case 11:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+304)%7;break;
    case 12:w=(365*(y-1)+((y-1)/4-(y-1)/100+(y-1)/400)+d+334)%7;
    }
    switch(w)
    {
    case 1:printf("1\n");break;
case 2:printf("2\n");break;
case 3:printf("3\n");break;
case 4:printf("4\n");break;
case 5:printf("5\n");break;
case 6:printf("6\n");break;
case 0:printf("7\n");
    }
}