Initially the order of problems was A-C-E-D-B. But we were not sure about last two.

This is simple straight-forward problem — you were asked to sort the teams with the following comparator: (*p*_{1} > *p*_{2}) or (*p*_{1} = *p*_{2} and *t*_{1} < *t*_{2}). After that you can split the teams into groups with equal results and find the group which shares the *k*-th place. Many coders for some reason used wrong comparator: they sorted teams with equal number of problems by descending of time. Such submits accidentally passed pretests but get WA #13.

Polygon A is convex, so it is sufficient to check only that every vertex of polygon B is strictly inside polygon A. In theory the simplest solution is building common convex hull of both polygons. You need to check that no vertex of polygon B belongs to this hull. But there is a tricky detail: if there are many points lying on the same side of convex hull than your convex hull must contain all these points as vertices. So this solution is harder to implement and has some corner case.

Another solution: cut polygon A into triangles (by vertex numbers): (1, 2, 3), (1, 3, 4), (1, 4, 5), ..., (1, *n* - 1, *n*). The sequences of angles 2 - 1 - 3, 2 - 1 - 4, 2 - 1 - 5, ..., 2 - 1 - *n* is increasing. It means that you can find for each vertex of B to which triangle of A it can belong using binsearch by angle.

Similarly you can cut polygon A into trapezoids (with vertical cuts). In this case you'll need a binsearch by *x*-coordinate.

If the initial array doesn't contain number *x*, than you definitely need to add it (that's +1 to answer). Than do the following. While median is strictly less than *x* you need to increase it. Obviously the surest way to increase the median is to add a maximal possible number (10^{5}). Similarly while the median is strictly more than *x*, add a number 1 to the array. Constraints are small, so you can add the numbers one by one and recalculate the median after every addition.

Also there is a solution without any cases: while the median isn't equal to *x*, just add one more number *x* to array.

Let's sort the people by decreasing of shoes size. Observe that when considering the *i*-th man we are interested in no more than 2 pairs of shoes: with size *l*_{i} and *l*_{i} + 1. It allows solving with dynamics. The state will be (the number of first unconsidered man *i*, is pair of shoes with size *l*_{i} available, is pair of shoes with size *l*_{i} + 1 available). You have 3 options: leave the *i*-th man without shoes or sell him a pair of shoes of one of suitable size (if available).

Obvious solution with dynamics: you need to know only how many moves are left and where is the ant. This is 4*n* states, each with 3 options – most of such solution passes. Observe that the vertices A, B, C are equivalent. This allows writing such solution:

```
int zD = 1;
int zABC = 0;
for (int i = 1; i <= n; i++) {
int nzD = zABC * 3LL % MOD;
int nzABC = (zABC * 2LL + zD) % MOD;
zD = nzD;
zABC = nzABC;
}
cout << zD;
```

Also this problem could be solved by *log*(*n*) with binary exponentiation of some 2 × 2 matrix into power *n*.

Edited -- Fixed my code! Thanks to Igel_SK!

Isn't the state for D too big?

Also, I see people doing bipartite matching in a greedy way, why does this work?

Bipartite matching with sorting shoes works correctly, it can be proofed (in russian). I think it should get TLE but the tests were weak :)

I solved with the DP this way: you have the current person and a mask that represents if the 2 sizes of shoes that this person can use are available (costumers are sorted by shoe size), so it will be a dp[10⁵][4] table.

Great explainatiions:) Can you explain your second solution on Problem B in detail ? Thx How do you perform binsearch on angles to calculate whether every vertex is in Polygon A or not

I solved E:Tetrahedron using the problem (3^n+3*(-1)^n)/4 . Here is my solution . However it failed the system test and when I used BigInteger it passed.Here is the 2nd version. 2 question: 1.Why does the first solution fail... 2.Is there any thing I can do to avoid BigIntegers

Use

`long`

Doubles are not accurate

Your code with long instead of double 1403711

Got the same problem.

Hi , can you explain your solution for this question some better ? I mean i know the your solution is true but how did you reach this solution ? Plz answer

how did you arrive at this formula can you please explain??

do the adjacency matrix A (in this problem you will have a matrix where A[i][j] = 1 if i!=j and A[i][j] = 0 if i==j). now A^2 count paths with two edges, A^3 count paths with three edges and so on, do exponentiation using diagonalization i.e A = P*D*P^(-1)

A^n = P*D^n*P^(-1)

You have to find a formula for one element on the diagonal using matrix P, D^n and P^(-1), just there you will notice the formula (3^n+3*(-1)^n)/4. (all of elements on diagonal are equal for this reason you need to find the formula of only one of them)

be carefull doing division using modulo, here you will find good information of how to do that https://codeaccepted.wordpress.com/2014/02/15/output-the-answer-modulo-109-7/

Hey! I am a beginner learning coding. I was solving 166E in problem set but couldn't. Read your comment and solution can you please explain me your code after ans%=mod on line 34? I will be thankful to you.

What 2 x 2 matrix can be used in E? I've only seen it solved with a 4 x 4 matrix with diagonal = 0 and rest = 1.

See Egor's solution: 1390234

Can you explain the idea behind this solution?

In general, you can do matrix exponentiation (that is, raising matrix M to the N-th power) in

`O(logN)`

time, where N is the exponent. If you exploit a "matrix notation" for your dp recurrence, then you can reduce it to this problem, thus solving it in`O(logN)`

.For example, it is possible to find the N-th fibonacci number using this method. For a better explanation, check out the "Fibonacci log N" tutorial on e-maxx.ru

let A be a matrix of 4X4 where,

`A[i][j] = 1 if i != j, and 0 otherwise`

Now, A represents the number of paths from i to j using 1 edge. Using some basic intuitions we can prove that A*A will represent the same but using 2 edges. So, we can A^n will represent all the paths from i to j using n number of edges. Now, you can exponent it or derive a formula. I derive a formula, see it here 51966083

Sorry about that, just ignore the post.

from math import *

def fast_exponentiation(base, n): if n == 1: return base else: if n % 2 == 0: base1 = fast_exponentiation(base, n/2)**2 return base1%1000000007 else: return (base*fast_exponentiation(base,(n-1)/2)**2)%1000000007

t = int(raw_input()) if t%2 == 0: print ((fast_exponentiation(3,t) + 3)/4)%1000000007 else: print ((fast_exponentiation(3,t) — 3)/4)%1000000007

Why is this code in Python wrong for thetraedron?

Note: you can use the built-in function

`pow`

for fast modular exponentiation:I keep getting wrong answer...

I get right values if I put modulus outside all the expression, so:

print ((fast_exponentiation(3,t) + 3)/4)%1000000007 gives correct value, but slow and print ((fast_exponentiation(3,t) + 3)/4) within the fast exp with mod gives WA, why?

You can't divide by 4 if you use mod before it. You should multuply pow(4, p-2, p) instead

Try ((fast_exponentiation(3,t) + 3) * 250000002) % 1000000007, and take modulo on each fast_exponentiation step.

ant.ermilov thank you for the tip, I have seen it in a solution as well, but can you explain to me, or point me to somewhere that can tell me, why is that particular number used?

At first, let's try to understand, why your approach does not work. For example, we have

Y = M+1, Y % M = 1,but

(Y / 4) % M != (Y % M) / 4.So division can be substituted by multiplication. In general, we have modulo M, divisor D and want to find such Z that for each X: (X / D) % M = ((X % M) * Z) % M.

After simple transformation we will get (X * Z) % M = 1.

For particular case 250000002 * 4 = 10

^{9}+ 7 + 1.Some extra info: Z can be found iff gcd(X, D) = 1, and simpliest way described at AlexDmitriev's comment above. From Euler's theorem we get Z = (

D^{M - 2}) % M.Thank you very much to both :D I got axcepted :)

Amazing explanation! Thanks

What does

non-degeneratemean in 166B - Polygons (Div2 B) ?Found myself : link

@AlexDmitriev Can you please explain In the solution to E why has nzABC been multiplied by 2 in the fifth line?

`int nzABC = (zABC * 2LL + zD) % MOD;`

Thanks a lot.

Let

A_{i}denote number of ways to finish near the vertexAafterimoves (same forB_{i},C_{i},D_{i}). Easily,A_{i}=D_{i - 1}+B_{i - 1}+C_{i - 1}B_{i}=D_{i - 1}+A_{i - 1}+C_{i - 1}C_{i}=D_{i - 1}+A_{i - 1}+B_{i - 1}D_{i}=A_{i - 1}+B_{i - 1}+C_{i - 1}with the initial conditions

A_{0}= 0,B_{0}= 0,C_{0}= 0,D_{0}= 1Due to the symmetry $A_i=B_i=C_i=ABC_i$, so

ABC_{i}=D_{i - 1}+ABC_{i - 1}+ABC_{i - 1}=D_{i - 1}+ 2 *ABC_{i - 1}D_{i}=ABC_{i - 1}+ABC_{i - 1}+ABC_{i - 1}= 3 *ABC_{i - 1}Thanks a lot!!

Nice Explanation :)

A path like A-B-D-C-D is valid, is it?

yes

thanks a lot

_meshanya_ How do you know that points A, B, C are symmetric? What do mean by symmetric? Geometric symmetry or some other symmetry?

Great explanation Indeed!

TETRAHEDRON : can this problem be think in a way that if u want to reach at D in n steps, then to travel n — 1 steps 3 ^ (n — 1) ways possible, now this case include that if u reached at D in n — 1 steps therefore to reach in n steps ans(in n steps) = 3 ^ (n — 1) — ans(in n — 1 steps); which is a DP solution

eg — to reach in 1 steps = 0 to reach in 2 steps — 3 ^ (2 — 1) — 0; to reach in 3 steps — 3 ^ (3 — 1) — 3 = 6 to reach in 4 steps — 3 ^ (4 — 1) — 6 = 21 to reach in 5 steps — 3 ^ (5 — 1) — 21 = 60

Thanks for the explanation.

Nice Explanation thank you

beautiful explanation.

I'm interested in E,whether there is any simple formula without matrix.i can't get it,anybody got it and coubld you share it?

CAN anybody explain E in detail? Can't get it.Thanks!

Let

f[A][j] is the number of ways from D to A by going j steps,as same,f[B][j],f[C][j] andf[D][j], we knowf[D][0] = 1,f[A,B,C][0] = 0(because the ant starts at D).then you get four equations below:

f[A][j] =f[B][j- 1] +f[C][j- 1] +f[D][j- 1]f[B][j] =f[A][j- 1] +f[C][j- 1] +f[D][j- 1]......

f[D][j] = ...you can solve this problem in

O(n),the answer is .but it's too slow.you can express the four equations by matrix.then it's obvious to solve the problem by Successive Square Method with matrix in .

@RAD In question B Polygons, What's the problem if we directly compute the convex hull of all points and check if any point of B is present in it ? I couldn't understand the mistake!!

Hi... . Could you please explain why is the output for testcase #13 of problem 166A - Rank List 1?

I am getting a RTE when I am submitting the solution for A. I have used a custom sort function and am baffled as to why I am getting the RTE sometimes (it is giving RTE sometimes only). Here is the code

35896046

Thanks in advance!

My

Python 3.6solution fails withO(n)complexity.I was able to solve Div 2 C Median using an O( 1 ) approach. Submission is here: https://codeforces.com/problemset/submission/166/52647089 .

Basic Idea: We have to ensure that the element we have to make as median (x) has to be at the central position. Now there can be 3 types of elements obviously: (i) smaller than x (ii) equal to x (iii) larger than x

Note that there can be duplicates so the number of elements equal to x can be more than 1 (or even 0). Our final aim is to ensure one of the elements equal to x lands up at the central position at the end. We know that in the final answer, 0 <= number of elements on right of median — number of elements on left <= 1 In case there are more than one equal elements, we can observe that the optimal answer will always have more of the extra equal elements on the side with less number of elements ( smaller or larger ). And the situation will look like this:

Now we have to find the number of elements that need to be added: Case 1: When number of smaller elements < number of larger As the left side can have one lesser elements than the right, number of elements we need to add is (elements greater than median — elements less than median — (elements equal — 1) — 1). Why elements equal — 1 ? Because one of these elements shall be at the median position. Obviously, it can't be less than 0, so we have to max it with 0. And also we subtract an additional -1 as we are allowed to have left side smaller.

Similarly we can devise Case 2 and 3 where elements on both side are qual or smaller > larger. I guess code will be able to explain itself better now !

I can't get the E problem n = 4; D-A-B-C-D (3!) D-A-D-B-D (2!) D-B-D-C-D (2!) D-A-D-C-D (2!) D-A-D-A-D D-B-D-B-D D-C-D-C-D That's 15 not 21 or I missed something?