T1
A math problem, understanding of the meaning of the questions on the test are some problems, I still have not carefully studied, so the pigeons
T2
Was first determine whether there is a legitimate program, if there is a legitimate program fell inquire title every $ x $, and push down the process somewhat like simulation, let's discuss how to determine whether there is a legitimate program, which used a DP you dare? $ skyh $ examination room AC, ye people think, we do not know, I do not dare to ask
Legal judgment
Provided $ dp [i] [j] $ denotes performed after the first $ $ I op, whether there may be 1, $ dp [i] [j] = 0 $ represents no solution, $ dp [i] [j] Representative = 1 $ solvable, then what we consider to $ dp [i-1] [j] $ can be transferred to the $ dp [i] [k] $ in, $ K $ what conditions need to meet
For $ and $ operation, our original result there have been a number $ j $, and the current can also be a $ a [i] $ one-number, the number that we get the maximum of $ { \ min} (a [i], j) $ th a, should have at least $ \ max (0, a [i] + jm) $ th a front half of a good understanding, completely overlap, the second half has to be coincide
For $ or $ operation, the minimum number obtained $ \ max (a [i], j) $ th a, up to $ \ min (m, a [i] + j) $ th a
For $ xor $, a minimum of a situation which had all the other numbers are offset by a number in this case have $ | a [i] -j | $ a one, then one is all stagger, and that this when there $ \ min (m, a [i] + j) $ th one, but obviously there is another constraint is the number, $ $ XOR result is a zero must be a 0 and a 1 $ XOR $ come, and so it needs $ 2 * ma [i] -j $ taken at $ {\ min} $
Looking for solutions
Set up a $ c $ which has $ num $ months, then $ dp [n] [num] $ If equal to 1, we will be able to find a combination of law program, then we discuss the process backwards
For XOR $ $, assuming $ y $ $ xor $ $ x = c $, $ c $ and we known in a number of $ $ XY, XY seeking $ $, due to satisfy the constraints for a number, we first thought should be in place in the $ c $ 0 to $ xy $ in both a put, in fact, there are several places to put the needs of a 0, can be counted out, there should be $ \ frac {a [ i] + j-num} {2} $, so we find to be more than $ $ XY put a whole, and then the remaining places for $ c $ $ X $ 1 to $ Y $ or put a can
$ And $ and $ or $ simpler, you can look at yourself yy
T3
In other positive solution seems to have root DP and the like, did not look carefully, it has the Gugu Gu