39 analog solution to a problem

A. Industrial title

Hand to play the sample.

And then found that the number of programs to reach the point above points are asking about take a, b of power.

A simple combination of several problems.

 

 

 

B. Chang title card

A cost flow, but only 40 points.

Test after the discovery cost flow is wrong, do not build map.

It may form part of a sub-prompt macrocycle.

Points 2n, 2n edges,

If not, what will form a large ring? Obviously ring tree.

Casual trees dp ring on the end, however, this fight is very troublesome.

In fact, the right part of the point can be seen as a side, still ring tree.

But the problem for conversion must choose a side next to each point, which is much simpler.

 

 

 

C. metaphysical questions

Crazy Mobius inversion exams.

Then get a $ O (m + n \ sqrt n) algorithm, and a violent line $ sieve fraction.

In fact, we are concerned only $ f (x) = \ sum \ limits_ {j = 1} ^ {m} Parity d (x * j) $ a.

Consider $ d (x * j) of $ parity, the number of simple expressions factor decomposition of the quality factor.

Obtained only if and when the number of $ x * j $ all power factor is even $ d (x * j) $ is an odd number, the answer to contribute.

Think of this exam, then I did not want to go.

An even total number of power factor, may be expressed as equivalent to a perfect square.

Thus the process of how many 1 ~ n each make up at least by a perfect square, set $ $ P_i

To enumerate later using 1 ~ m, $ f (i) is the parity $ $ \ sqrt \ frac {m} {p_i} $ parity.

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Origin www.cnblogs.com/skyh/p/11482558.html