No problem was specially taken from "Wikipedia". For me it's really hard to understand, that problem "B" was there.

**A (Div 2).** (Idea : Gerald, realization : Aksenov239, code : 1670635)

If the lengths of 2 strings aren't equal — that means "NO". We try to find the positions in strings, where chars are different. If there 1 or more than 2 such positions — "NO". After that we swap 2 characters in the first string, and check for their equality.

**B (Div 2).** (Idea : Aksenov239, realization : Aksenov239, code : 1670637)

- We can see, that we can do all in integers, because k is integer number of percent.
- For each dwarf we should find his optimal strategy — to check 2 strategies with speed.
- We should sort them.

**A (Div 1).** (Idea : Aksenov239, realization : Aksenov239, code : 1652592) Let's propose, that after the *i*-th year, there is *x* triangles up and *y* triangles down. After another iteration we can see, that amount of triangles became — 3*x* + *y* up and *x* + 3*y* down. Let's see the difference between them: at the *i*-th it's *x* - *y* and at the *i* + 1-th — it's (3*x* + *y*) - (*x* + 3*y*) = 2 * (*x* - *y*). We can see, that difference between amount of triangles grown up by 2. Because on the *i*-th year the difference became 2^{i} and all amount of triangles is 4^{i}. We can see, that on the *i*-th year the number of our triangles is . That can be computed by modulo p using the fast-power algorithm.

**B (Div 1).** (Idea : Aksenov239, realization : Aksenov239, code : 1652597) This problem was made by my love to inequalities. :-)! The answer for this problem is .

Prove: . (This is AM-GM inequality. Link for whom don't know it.)

The equality becomes only, when .

And you should check on zeroes. If *a* = *b* = *c* = 0 — you can choose any good answer — *x* + *y* + *z* ≤ *S*.

**C (Div 1).** No comments. :-)!

**D (Div 1).** (Idea : Aksenov239, realization : cerealguy, Gerald, Aksenov239, code : 1652604)

This is number theory problem.

I'm trying to explain it step by step:

1) Let's prove, that LCD is maximum 2.

Let . Squaring both sides we get , but we want to . This means, that *d* can be only 2.

2) Let's make this lenghty product simplier.

.

We can count this by modulo *p* fast, and divide it by 2^{r - l}, if *k* is odd.

3) There is many interesting things in this solution.

Firstly, it doesn't work, when *p* = 2 — but it can easily done by you.

The other problem is harder, what if , this means that for each *i* ≥ *l* : , and this mean,

that for each *i* ≥ *l* : *k*^{2i} + 1 ≡ _{p}2. And the product by modulo *p* is equal to 2^{r - l + 1}.

E (Div. 1) (Idea : Aksenov239, relization : cerealguy, code : 1652611)

This problem wasn't taken to ROI, because of that I gave it here. This is pretty hard problem. I can't now realize, what cerealguy wrote, but his solution is *O*(*nlogn*) — without binary search. For me it's quite hard to understand, because my first solution was with binary search. And there were solutions, that has a worse asymptothic, but they run faster.

Because of that I can only give you key ideas, that can help you. (afterwards you can see the code of cerealguy)

Ideas:

- Let's find for each person the nearest subway point for them. It can be done in nlogn with use of segment tree or something else.
- We can see, that if one person goes to subway, the others, which distance to subway is smaller, can go to subway too — it doesn't affect the answer. Because of that we sort all persons by their distanse to subway.
- The area of the person, where he can come in
*t*seconds, is romb. And we can intersect all rombes in*O*(*n*) — the intersection is like rectangle. - Let's make the first intersection. When nobody goes to subway. We get a rectangle.
- The main idea, that for this rectangle — the nearest subway becomes always the same. We go throught people in sorted order, we can fast recalculate this small rectangle — and because of that we can fast recalculate the answer.

And we get a solution in *O*(*nlogn*) time. Ta-dams.

Thank you all for your attention. I'm deeply sorry about the problem "C".

For some time links for solutions isn't working. Temporarily:

А div. 2 — http://pastie.org/3884140

B div. 2 — http://pastie.org/3884143

A div. 1 — http://pastie.org/3884145

B div. 1 — http://pastie.org/3884147

D div. 1 — http://pastie.org/3884149

E div. 1 — http://pastie.org/3884152

great tutoriAL!!!!!!!!!!

I just think cf need to save the draft as I accidently click on some link and the content is lost……

I may need to borrow a book about number theory……

and at last……I just can't understand the product in C(Div.1)

a rookie

Try this:

Multiply LHS by "k^(2^l) — 1" and get: (k^(2^l) — 1) (k^(2^l) + 1) ... (k^(2^r) +1)

Repeatedly apply the identity: (a — b)(a + b) = a^2 — b^2 (for every first two terms)

(I accdiently posted in russian and I can't see my comment, I'm sorry for people who see this comment twice)

I did the math for div 1 B and got: x = (aB/(a+b+c))^(a+b+c) y = (bB/(a+b+c))^(a+b+c) z = (cB/(a+b+c))^(a+b+c)

I cant find the movtivation of why we get rid of the power to get the real answer.

A (Div 1) : simple explain

suppose : up -> num of upward triangle and down -> num of downward triangle

Initially :

conculsion : each n we split each triangle to 4 triangles it's mean at n we have 4^n triangle we want to want how many of 4^n are upward

suppose : a -> num of upward triangle b -> num of downward triangle

fact :

conculsion :

also we see that sum of a' + b' = [ (2^n)+1 ] + [ (2^n)-1 ] ,

from (2) : a' = [ (2^n)+1 ] , b' = [ (2^n)-1 ]

[ a' + b' ] * c = 4^n

[ 2^(n+1) ] * c = 4^n

I solved Div1 B using Lagrange's Multiplier .

L(x,y,z,lamda)=x^a*y^b*z^c-lamda*(x+y+z-s) and then partially differentiating w.r.t x,y,z and equating to 0 gives x=a*s/(a+b+c) y=b*s/(a+b+c) z=c*s/(a+b+c)