you all know that creat2012 will go to a regional. and i want you all pray for us. thx.

  for this problem, you need judge if two segments is intersection(相交的). if intersect(相交), print yes and the point. else print no.

  easy ? ac it.

 T <= 100 cases
8 real numbers (double)
promise two segments is no coincidence.

 no
or yes and two real number (one decimal)

 不会出现线段重合的情况。

黄庆浩