SRM 553

p250: 0 is special, so eval with program[pos] set to 0, where pos is the -1.
then you can eval with -1 there. If it comes out equal wanted, then you return 1, since 0 did not work and the next smallest number is 1.
Otherwise you got a number different from wanted. You know that if target can make eval to wanted, then use -1 instead of target you get wanted - target - 1, so target = wanted - ans -1, where ans is the eval result when you leave -1 unchanged. However there is a catch, the eval returned value may not use -1 at all. To trace whether -1 has been hit or not turns out a bit tricky. pieguy has an elegant solution, use 1 and 2 to replace -1 and eval them, if the two ans come out different, then you know -1 has been hit. My offline solution is to do one more eval starting from program[pos]=-1, then check stack size = 1. There are other solutions which remembers whether the current value in stack has -1 built in or not.

p500, the solution must be a decreasing rows of B matched with an increasing rows of W, this will lead to a dp solution. One more observation, since each cell is either B or W, it must be start with topleft an all B or all W row echelon form.