Topcoder SRM 772 - Codeforces

5 лет назад, # |

← Rev. 3 →

+12

500 is a really nice problem!

I expect somebody to say "Oh, it was in Polish subregional in Chrząszczyżewoszyce in 2012, it's such a standard trick". For me, it was fresh.

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5 лет назад, # ^ |

thank you :)

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yassin_

5 лет назад, # |

Great, can't even solve 250... The combinatorial formula seems obvious, but how to compute it fast since M is non-prime??

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5 лет назад, # ^ |

+13

you can simplify it to get answer as a product of contiguous integers which can be computed using segment tree or any data structure.

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5 лет назад, # ^ |

+16

The answer is $$$\left[X \cdot (X-1) \cdot \dots \cdot (X-K+1)\right] \cdot (N-K)!$$$. You can compute the first term using Fenwick tree with multiplication modulo $$$M$$$.

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nitesh_gupta

5 лет назад, # ^ |

+25

How can we use Fenwick tree to calculate the first term? Won't we need to calculate multiplicative inverse in that case?

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5 лет назад, # ^ |

Observe how you calculate such a product with a segment tree. In each node, you either

return the value stored in the node if its interval is subset of the query,
return zero element if its interval is disjoint with query, or
recurse into both children and combine the result

You can do the same with bitwise magic in a Fenwick tree.

I don't have that in my codebook, so I simply used a (slightly slower) segment tree with multiplication over a finite field:

Code

MOD = M;
auto F = RField::fact(MX+1);

vector<RField> W(MX+1);
for (int i = 0; i <= MX; ++i) W[i] = RField{i};
SetMulTree<RField> ST;
ST.setup(W, RField{0});

ll tot = 0;
for (int i = 0; i < T; ++i) {
    if (K[i] > X[i]) continue;
    RField ans = ST.get(X[i]-K[i]+1, X[i]) * F[N[i]-K[i]];
    tot += (int)ans;
}
        
return tot;

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aryanc403

5 лет назад, # ^ |

← Rev. 2 →

+14

Binary Lifting?

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neal

5 лет назад, # ^ |

+23

Here's an image of a Fenwick tree.

How would you compute the product of the range [4, 7]? Unlike in a full seg tree, there are no segments starting at 4.

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5 лет назад, # ^ |

Oops, you're right. I've read about a modification of Fenwick tree to support min-queries, and mixed it up.

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vlyubin

5 лет назад, # ^ |

Even a slightly easier option is to use sqrt-decomposition. You need to add 4 lines of code to the loop computing the result.

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square1001

5 лет назад, # |

Really nice round! For Div1 all three were nice problems, especially for Div1 500 — it was so moving when I have come up with the solution. Definitely one of the best problems of this year!

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5 лет назад, # ^ |

+22

thank you a lot!! When I thought of this problem, the first solution which came to my mind was using max flow, then when I observed the flow carefully I realized that it could be done using greedy strategy.

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ksun48

5 лет назад, # |

+46

1000:

The terms you need to sum have the form $$$\frac{N!}{R! (N-R)!} \cdot \frac{1}{R + Q} \cdot P(R)$$$, summed from R = 0 to N.

The $$$R+Q$$$ term is annoying, but if we absorb it into the $$$R!$$$ term, we can rewrite this as $$$\frac{N! (R+1)\dots(R+Q-1)}{(R+Q)!(N-R)!} P(R)$$$, which can be seen as $$$\binom{N+Q}{R+Q}$$$ times some new polynomial in R, $$$P_2$$$. Polynomial $$$P_2$$$ has small degree as K and Q are small.

Now this can be computed by the binomial theorem: $$$\sum_{i=0}^{A} \binom{A}{i} (i)(i-1)\dots(i-k+1) = A(A-1)\dots(A-k+1)\sum_{i=0}^{A-k} \binom{A-k}{i} = A(A-1)\dots(A-k+1)2^{A-k}$$$. It's left to represent $$$P_2$$$ as a sum of polynomials of the form $$$x(x-1)\dots(x-k)$$$, where k ranges between [Q-1, Q+K-1].

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5 лет назад, # ^ |

-10

We can also kinda get rid of the annoying $$$R+Q$$$ by getting rid of everything else instead. Just write $$$R = (R+Q)-Q$$$, $$$R+P = (R+Q)+(P-Q)$$$ and expand using the binomial theorem, which leaves us with $$$\sum_{-1 \le k \le K} \sum_R {N \choose R} (R+Q)^k A_k$$$. The terms with non-negative $$$k$$$ can again be expanded and $$$S_{N, k} = \sum_R {N\choose R} R^k$$$ can be rewritten as e.g. matrix exponentiation.

Then we need to compute $$$\sum_R {N\choose R} \frac{1}{R+Q}$$$, which seems to have a reasonably nice closed form (says WolframAlpha for small $$$Q$$$).

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ksun48

5 лет назад, # ^ |

← Rev. 2 →

What's the closed form? (To me, computing this sum is the hard part of the problem.)

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5 лет назад, # ^ |

← Rev. 5 →

The answer to that can be obtained by integrating x^(Q-1)*(1+x)^N under limits 0 to 1 which can be done using dp in O(Q) time. :)

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5 лет назад, # ^ |

for Q=4, the denominator is clear and the numerator is 2^n * poly(n) + constant...

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ksun48

5 лет назад, # ^ |

Sure, but what's the polynomial, if you say there's a closed form? Even if you have an expression for the polynomial, it has degree Q, and doesn't necessarily follow that you can easily evaluate in O(Q) time instead of O(Q^2).

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5 лет назад, # ^ |

I said seems to have, it looks nice enough that it could have one. Maybe it doesn't.

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hos.lyric

5 лет назад, # ^ |

+10

Actually when $$$Q$$$ is fixed it is P-recursive of order $$$2$$$ and degree $$$1$$$, so the values for $$$N = 0, ..., Q$$$ can be computed in time $$$O(Q)$$$.

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Petr

5 лет назад, # ^ |

← Rev. 2 →

+13

My thoughts without Wolfram Alpha: consider the polynomial f(x)=sum(c(n,r)*x^(r+q)/(r+q)). We need to find f(1). f'(x)=sum(c(n,r)*x^(r+q-1))=x^(q-1)*(1+x)^n. Now we need to find the integral of that, which we can do by integrating by parts q-1 times. The formula will thus have q terms with some powers of two and partial factorials.

Didn't manage to code this in time, though :)

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5 лет назад, # ^ |

+11

Good idea. It's indefinite $$$\int (x+c)^a (x+d)^b = \sum_{k=0}^b \frac{(-1)^k (x+c)^{a+1+k} (x+d)^{b-k} a! b!}{(a+1+k)!(b-k)!}$$$, so

$$$\sum_r {N \choose r} \frac{1}{r+Q} = \sum_{k=0}^{Q-1} \frac{(-1)^k 2^{N+1+k} N! (Q-1)!}{(N+1+k)!(Q-1-k)!} + \frac{(-1)^Q N! (Q-1)!}{(N+Q)!}$$$

and then, $$$\sum_r {N \choose r} r^a$$$ can be computed as $$$\sum_r {N \choose r} r^a e^{rx}$$$ at $$$x=0$$$, which is the $$$a$$$-th derivative of $$$\sum_r {N \choose r} e^{rx} = (1 + e^x)^N = (1 + \sum_{k \ge 0} \frac{x^k}{k!})^N$$$, which is $$$a!$$$ times the coefficient at $$$x^a$$$ of this sum. Since $$$a$$$ is $$$O(K)$$$ here, this can be found easily with fast exponentiation.

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5 лет назад, # ^ |

← Rev. 2 →

Can you explain how do you calculate $$$a$$$ th derivative of $$$(1+e^x)^N$$$ .

I couldn't understand this part: "which is $$$a!$$$ times the coefficient at $$$x^a$$$ of this sum. Since $$$a$$$ is $$$O(K)$$$ here, this can be found easily with fast exponentiation.".

Edit: Got it. Nice approach :)

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hos.lyric

5 лет назад, # ^ |

Here goes my ugly solution (which I completed in during the challenge phase :(

$$$\sum_{r=0}^N \binom{N}{r} r^k = 2^N f_k(N)$$$ where $$$f_k$$$ is a polynomial of degree around $$$k$$$.

$$$\sum_{r=0}^N \binom{N}{r} \frac{1}{r+Q} = \frac{(Q-1)!}{(N+1)\cdots(N+Q)} (2^N g_Q(N) + (-1)^Q)$$$ where $$$g_Q$$$ is a polynomial of degree around $$$Q$$$. I found this by looking at the sequence, making them integers, and using OEIS...

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Benq

5 лет назад, # ^ |

← Rev. 2 →

You need to compute sums of the form

$$$\sum_{i=0}^{N+Q}\binom{N+Q}{i}(i-1)(i-2)\ldots (i-k)$$$

for $$$Q-1\le k\le Q-1+K,$$$ but it isn't immediately obvious how your explanation above addresses this. It took me a while to realize that this sum can be easily computed given

$$$\sum_{i=0}^{N+Q}\binom{N+Q}{i}i(i-1)(i-2)\ldots (i-k)$$$

and

$$$\sum_{i=0}^{N+Q}\binom{N+Q}{i}(i-1)(i-2)\ldots (i-k+1)$$$

...

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SpoonsAN

5 лет назад, # |

How to solve div1 — 500 ?

I can only solve the specific case which all the points are arranged in a line. It can be solved with a greedy algorithm, taking the points one by one from the two sides of the line to the "outside set".

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