chokudai's blog

AtCoder Beginner Contest 156 Announcement
By chokudai, history, 5 years ago, In English

We will hold AtCoder Beginner Contest 156.

The point values will be 100-200-300-400-500-600.

We are looking forward to your participation!

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5 years ago, # |
← Rev. 2   Vote: I like it -10 Vote: I do not like it

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    5 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Positional notation (or place-value notation, or positional numeral system) denotes usually the extension to any base of the Hindu–Arabic numeral system (or decimal system). More generally, a positional system is a numeral system in which the contribution of a digit to the value of a number is the product of the value of the digit by a factor determined by the position of the digit. In early numeral systems, such as Roman numerals, a digit has only one value: I means one, X means ten and C a hundred (however, the value may be negated if placed before another digit). In modern positional systems, such as the decimal system, the position of the digit means that its value must be multiplied by some value: in 555, the three identical symbols represent five hundreds, five tens, and five units, respectively, due to their different positions in the digit string.

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    5 years ago, # ^ |
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    Quoted Wikipedia

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5 years ago, # |
  Vote: I like it +25 Vote: I do not like it

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5 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Problem E, my idea:

r1 + r2 + ... + rn = n , such that 0<=ri<=min(k+1,n) (ri is the no. of people in that room) ,

what's wrong with this ?

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    5 years ago, # ^ |
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    There is more than one way to choose from the people on initial positions. It should fail on the 3rd sample case.

    The correct solution is when you choose how many will be 0 and then you do that splitting. C(N, X) * StarsAndBars(X, N-X);

    (0<=X<=min(N-1,K))

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    5 years ago, # ^ |
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    n = 3 and k = 1

    1 1 1 has 0 <= ri <= min(k+1 , n), but its an invalid configuration

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    5 years ago, # ^ |
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    Think of this case : n=4, k=1.

    You counts this placement: [2 2 0 0] and its 5 other permutations.

    But to make this condition there should be at least 2 moves.

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5 years ago, # |
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how to solve E ??

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5 years ago, # |
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how to solve F?

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    5 years ago, # ^ |
      Vote: I like it +21 Vote: I do not like it

    Let x = x%m, d(i) = (d(i)%m!=0)?d(i)%m:m. Note that a(i)%m >= a(i+1)%m if a(i)%m + d(i) >=m. You would have to find the count of these which is nothing but a(n-1)/m. The answer to the problem is n-1 — a(n-1)/m.

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5 years ago, # |
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My approach of D is to calculate all the combinations ((2^n)-1) — nCk(n,a) — nCk(n,b), but the results are wrong.

Is this apprach wrong?

This is my code:

https://atcoder.jp/contests/abc156/submissions/10292388

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5 years ago, # |
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How to solve E ??

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5 years ago, # |
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How to solve problem C (rally)

I'm taking a mean of array elements for point P but failed to get AC in all test cases.

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    5 years ago, # ^ |
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    just apply brute force , iterate for each x , 1<=x<=100 and find total points for each and take minimum of them

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    5 years ago, # ^ |
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    Since the array of values is small you can brute force the solution. Just try every one of the 100 possible indexes.

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    5 years ago, # ^ |
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    enum p:1 to 100 ，then min{ans}

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    5 years ago, # ^ |
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    just check for min(avg,avg+1) . The function's((x-x1)**2+(x-x2)**2) derivative is zero at only x=sum/n . https://atcoder.jp/contests/abc156/submissions/10305420

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      4 years ago, # ^ |
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      Why is the function f(x) = ((x-x1)^2 + (x-x2)^2) ? Shouldn't it be f(x) = (x-x1)^2 + (x-x2)^2 + (x-x3)^2 + .. + (x-xn)^2?

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5 years ago, # |
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How to calculate nCr%p, when n=10^9 and p=10^9+7? I got the formula for problem D(2^n-1-nCa-nCb), but couldn't solve it.

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    5 years ago, # ^ |
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    ncr can also be calculated in O(r) like ncr= (n*(n-1)*(n-2)*(n-3)......(n-r+1))/(1*2*3.......*r) so for numerator directly iterate and take modulo and for denominator take modular inverse of them

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    5 years ago, # ^ |
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    I have answered this previously, in this blog. You can have a look at it.

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5 years ago, # |
← Rev. 3   Vote: I like it +57 Vote: I do not like it

Quick solution sketches:

  • A: Implementation.
  • B: The number of base-$$$K$$$ digits in the representation of $$$N > 0$$$ is equal to 1 plus the number of digits in $$$\left \lfloor{\frac{N}{K}}\right \rfloor$$$, where $$$0$$$ is considered to have no digits.
  • C: The bounds are small enough to compute the total stamina spent for each choice of meeting point.
  • D: We are asked to compute $$$2^N - 1 - \binom{N}{a} - \binom{N}{b}$$$ modulo $$$10^9 + 7$$$, where $$$N \leq 10^9$$$ and $$$a,b \leq 2 \cdot 10^5$$$. We can compute $$$\binom{N}{a} = \frac{N!}{a!(N-a)!} = \frac{N \cdot (N-1) \cdot (N-2) \cdots (N-a+1)}{a!}$$$ quickly as both the numerator and denominator have only $$$a$$$ terms.
  • E: After moving people around, some rooms will decrease in population while others will increase in population. All rooms which decrease must contain no people. Suppose there are $$$x$$$ empty rooms. There are $$$\binom{N}{x}$$$ ways to select them. We can apply stars and bars to find that we have $$$\binom{N-x-1+x}{x}$$$ ways to distribute the people who left the empty rooms into the non-empty ones. Note that it is not possible for all rooms to be empty, or for more than $$$K$$$ rooms to be empty.
  • F: Before handling query $$$q$$$, compute $$$d'_i = d_i \mod m_q$$$. Instead of counting the number of times $$$a$$$ strictly increases modulo $$$m_q$$$, we'll count the number of times it strictly decreases and the number of times it stays the same. Each time we add some $$$d'_i \in [0, m_q)$$$ to produce a new element of sequence $$$a$$$, the value of $$$a$$$ modulo $$$m_q$$$ decreases iff we cross a multiple of $$$m_q$$$, and it stays the same iff $$$d'_i = 0$$$. We can compute the starting and ending values of $$$a$$$ to efficiently determine the number of multiples of $$$m_q$$$ we'll cross. We can also efficiently compute the number of $$$d'_i = 0$$$ occuring in the $$$n_q - 1$$$ terms of $$$d'$$$ which will be used.
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    5 years ago, # ^ |
    ← Rev. 2   Vote: I like it +13 Vote: I do not like it

    ( Related to Problem F ) Is there a quick way ( $$$O(1)$$$ or anything sublinear maybe ) to solve the following:

    Given integers $$$a,b,sum,m,ops$$$, find number of $$$j$$$ such that, $$$ ( (a + sum*j) \mod m ) < ( (b + sum*j) \mod m )$$$ , $$$0 <= j < ops$$$.

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      5 years ago, # ^ |
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      I don't think so, basically you have a linear congruential generator and you're asking how many of the first $$$ops$$$ terms fall in some interval.

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        5 years ago, # ^ |
          Vote: I like it +8 Vote: I do not like it

        In fact, it can be solved in $$$O(\log n)$$$. I asked my friends this question and skylinebaby told me that we can solve it by using the function $$$f (a,b,c,n)=\sum_{i=0}^{n} \lfloor \frac{a i + b}{c} \rfloor$$$

        $$$f()$$$ can be calculated in $$$O(\log n)$$$ by using some Chinese magic called 类欧几里德算法. (roughly translated as "Euclidean-like algorithm", the name comes from its similarity to Euclidean algorithm)

        Unfortunately, it seems that it's not fast enough to pass F. I tested it locally on testcase27 and it gave correct answer in 4 seconds. Check my code for details.

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          5 years ago, # ^ |
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          Wow. This was quite cool. Thanks for bringing this out. Looks like chinese have lots of magic.

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    5 years ago, # ^ |
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    In problem E why is it wrong to think of distributing the n people in the n-k non empty rooms. using the stars and bars we get (2*n+i-1) C n ??? thanks for the explanations.

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      5 years ago, # ^ |
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      I think there is a restriction that maximum people accumulated in any room is = K+1 and that too is possible with only one room, it cannot be satisfied if we apply stars and bars using n people and n-k empty rooms.

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    5 years ago, # ^ |
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    how (N-x-1+x)Cx ways are possible

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      5 years ago, # ^ |
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      Read about stars and bars.. Suppose you have to arrange n items in k boxes. We represent items with stars and boxes with bars. For n=10 and k=3 a situation could be like ...|..|..... That means 3 items in first box, 2 items in second box, 5 items in third box. The solution would be permutation of n+k-1 items with n similar items and k-1 similar items = $$${\frac{(n+k-1)!}{ (n)! * (k-1)! } }$$$

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    4 years ago, # ^ |
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    For problem E, should x always start from 0? or rather from 0 only when k is even and start from 1 when k is odd.

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5 years ago, # |
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Isn't problem C the same as this problem?

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    5 years ago, # ^ |
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    Yes, It's exactly the same.

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      5 years ago, # ^ |
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      As the problem title declares (「いっしょ」 means "the same" in Japanese).

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5 years ago, # |
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All problems solution with explanations by tmwilliamlin168. Link : https://www.youtube.com/watch?v=3fnkK4wPU14

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5 years ago, # |
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E was an interesting question. How to solve F?

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5 years ago, # |
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Just noticed that there is an english editorial available

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5 years ago, # |
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How to solve the Problem D?I've been thinking for a long time but i can't find a way to prevent time limit exceed.

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    5 years ago, # ^ |
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    $$${nCr}$$$ can ke solved in $$${O(k)}$$$ where k=r as $$${(n)!*(n-1)!*...*(n-r+1)!/(r)!}$$$

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5 years ago, # |
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I still don't understand the E......

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5 years ago, # |
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Where can i see failed test cases after contest in atcoder ?

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4 years ago, # |
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For those having difficulty in solving E.

I have explained my solution in my submission.

thank me later :)

https://atcoder.jp/contests/abc156/submissions/15300567

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