大夫值班问题：医院有A、B、C、D、E、F、G七位大夫，在一个星期内（星期一至星期天）
每天要轮流值班一天。现在已知：
A大夫比C大夫晚一天值班；
D大夫比E大夫晚两天值班；
B大夫比G大夫早三天值班；
F大夫的值班日在B和C大夫的中间，且是星期四；
请编程确定每天究竟哪位大夫值班？
**输出格式要求："Doctor %c is on duty %s.\n"
程序运行示例如下：
Doctor E is on duty MONDAY.
Doctor B is on duty TUESDAY.
Doctor D is on duty WEDNESDAY.
Doctor F is on duty THURDAY.
Doctor G is on duty FRIDAY.
Doctor C is on duty STAURDAY.
Doctor A is on duty SUNDAY.
答案如下，求大神解析一下思路：
#include
int a[8];
char *day[]={
"","MONDAY","TUESDAY","WEDNESDAY","THURDAY",
"FRIDAY","STAURDAY","SUNDAY"};
main()
{
int i,t,j;
a[4]=6;
for(i=1;i<=3;i++)
{
a[i]=2;
if(!a[i+3])
a[i+3]=7;
else
{
a[i]=0;
continue;
}
for(t=1;t<=3;t++)
{
if(!a[t])
a[t]=5;
else
continue;
if(!a[t+2])
a[t+2]=4;
else
{
a[t]=0;
continue;
}
for(j=5;j<7;j++)
{
if(!a[j])
a[j]=3;
else
continue;
if(!a[j+1])
a[j+1]=1;
else
{
a[j]=0;
continue;
}
for(i=1;i<=7;i++)
printf("Doctor %c is on duty %s.\n",'A'+a[i]-1,day[i]);
exit(0);
}
}
}
}
本来呢，就是要遍历所有情况，然后有种或者可能多种情况满足所有限制条件就是答案，然后根据限制条件提前去掉一些循环，比如ac捆绑的，就是这样。。。
一听说是妹子，我很想帮你，可我是玩c#的，所以，就来打打酱油咯