We will hold AtCoder Beginner Contest 156.

- Contest URL: https://atcoder.jp/contests/abc156
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20200222T2100&p1=248
- Duration: 100 minutes
- Number of Tasks: 6
- Writer: gazelle, kort0n, namonakiacc
- Rated range: ~ 1999

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

We are looking forward to your participation!

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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.

Quoted Wikipedia

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 ?

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))

n = 3 and k = 1

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

2<=k is stated in constraints.

$$$N = 6$$$, $$$K = 2$$$: you can't form $$$(3, 3, 0, 0, 0, 0)$$$

Yes, you can not.

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.

how to solve E ??

how to solve F?

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.

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

It is 2^n-1 — nCk(n,a) — nCk(n,b)

Sorry, i didn't mention that but i did subtract 1.

It's sad that Chinese can't visit https://ideone.com/

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

In line 35, you should use modular multiplicative inverse instead of dividing the number directly

Why it's wrong to divide directly?

Here is an example that you will get the wrong answer while dividing number directly under modulo.

For more detail about "modular multiplicative inverse": link

You can't do

`res /= (i + 1LL);`

directly. This is my code https://atcoder.jp/contests/abc156/submissions/10273339.https://atcoder.jp/contests/abc156/submissions/10285959 mod inverse is missing.

Why it's needed?

12%7=5

12%7/5=1

12/5%7=2（And this one certainly make no sense）

How to solve E ??

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.

just apply brute force , iterate for each x , 1<=x<=100 and find total points for each and take minimum of them

Since the array of values is small you can brute force the solution. Just try every one of the 100 possible indexes.

enum p:1 to 100 ，then min{ans}

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

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?

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.

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

Ok Thanks

I have answered this previously, in this blog. You can have a look at 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.( 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, $$$ ( (a + sum*j) \mod m ) < ( (b + sum*j) \mod m )$$$ , $$$0 <= j < ops$$$.

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.

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.

Wow. This was quite cool. Thanks for bringing this out. Looks like chinese have lots of magic.

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.

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.

how (N-x-1+x)Cx ways are possible

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 = $$${\frac{(n+k-1)!}{ (n)! * (k-1)! } }$$$

For problem E, should

xalways start from 0? or rather from 0 only when k is even and start from 1 when k is odd.Isn't problem C the same as this problem?

Yes, It's exactly the same.

As the problem title declares (「いっしょ」 means "the same" in Japanese).

All problems solution with explanations by tmwilliamlin168. Link : https://www.youtube.com/watch?v=3fnkK4wPU14

E was an interesting question. How to solve F?

Just noticed that there is an english editorial available

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.

$$${nCr}$$$ can ke solved in $$${O(k)}$$$ where k=r as $$${(n)!*(n-1)!*...*(n-r+1)!/(r)!}$$$

I still don't understand the E......

Where can i see failed test cases after contest in atcoder ?

https://codeforces.com/blog/entry/46389 ,see this blog

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

In Problem F.

if first term in sequence is X and last term is Y then, to count all terms for case : A[i]%m > A[i+1]%m

according to editorial number of them = (Y/m) — (X/m).

can someone explain how this formula can count number of them ?