Test 106 T1 20pts

T1:

Background: Simple question template
reasons: 1, the details wrong
     2, can not shoot
method:
1, the amount of output process, an alternative to the beat.
2, playing, check, want to run mode.

T2:

Divided into two parts, ab and final jump of about a. Referred to as X, Y.

X greedy big to small order.

Enumeration Y, n ^ 2 check

Optimization check process: If Y in the X, directly suffix maximum forward sweep sequence statistics.

Consider Y before X. X is regarded as the last pick out a leap of faith leap .     

I consider the contribution of L minus, plus a contribution of i's. lowerbound find a jumping off point, called p.

The 1 ~ i-1 must be legal, i ~ p regarded as a contribution to pan left. Plus one minus the i posterior translation.

W is defined as an array Sum [i + 1] -C [i]; i ~ pW ST table to find the minimum value, subtracting the contribution of i bits, see> to 0.

( Why do so many A randomized, ah, positive solution very reluctantly )

T3:

To achieve all-zero state:

a [i] [j] = 0, the position of the current point may limit the number of parity covering.

I actually row J column is selected or not selected at the same time or just one.

The lack of further examination room thinking, thinking about this when dfs, thought it was similar to the 2-sat but complicated holiday.

Transforming ideas: no longer dfs search program. Or divide the ranks of the opposition to the same collection. Dfs coloring into judgment.

To know can achieve all zeros.

Further thinking is successively hand.

A blind spot: a not link block, but several.

Discuss two opposing collection sz parity, divided into the same odd 11, with even 00, 10 are different.

Discussion can be.

More subtly sg function. Hand has transformed ideas.

sg (11) = 1, sg (00) = 0, SG (01) = 2;

And the exclusive OR is 0, B $ brother win. Otherwise, a small W M Z win.

Practical implications: (01) an even number flip equilibrium state, the upper hand odd selection status. (00) useless. (11) to win after the defeat.

      Only (11) even, (01) lost the upper hand when even.

      sg (01) = 2, different from (00), (11), because these are two domains, different roles.

Interpretations: sg (00) = 0.sg (11) take a step to the even block, sg (0) = 0, then sg (11) = 1;

sg (01) may be a parity, to 0/1, it is 2.

 

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Origin www.cnblogs.com/seamtn/p/11824302.html