The starting position can be anywhere with a footprint. The footprints can be categorized into 3 types.

- only
*L*s - only
*R*s *R*s followed by*L*s

In case 1, we end in the left of all footprints. In case 2, we end in the right of all footprints. In case 3, we either end in the rightmost *R* or the leftmost *L*

We can simply move by greedy method — only moves when it takes the boat closer to the destination.

**Fact 0: If a has odd parity, we can apply operation 1 to increase its number of 1 by 1.**

**Fact 1: If a has even parity, its number of 1 cannot increase anymore**

If *a* has parity 0, unless we pop a 1, otherwise we cannot write a new 1 into *a*.

**Fact 2: If the number of 1 in a is not less than the one in b, we can always turn a to b**

The idea is to make a copy of *b* at the right of *a*. Lets assume *a* has even parity now. If we need a 0, simply apply operation 1. If we need a 1, keep remove from head until we remove a 1. Notice that we never remove digits from 'new part' of *a*. Now the parity of *a* will be odd and we can apply operation 1. After that, the parity of *a* becomes even again, the number of 1 in the 'old part' of *a* decrease by 1 and we handle a 1 in *b*.

Finally, remove the remaining 'old part' of *a*. Now we get *b*.

Combine all those facts, we can conclude that we can turn *a* into *b* if and only if

*countOnes*(

*a*) +

*parity*(

*countOnes*(

*a*)) ≥

*countOnes*(

*b*)

If *n* > *m*, set every weight to 1 and we are done. Otherwise, sort *a* and *b* in non-increasing order, and trim the last part of *b* such that its length equals *a*.

**Claim: answer is YES if and only if exists i such that a_{i} > b_{i}**

If for every *i*, *a*_{i} ≤ *b*_{i}, that means for every Alice's fish, there is a corresponding Bob's fish which is as heavy as Alice's.

Let *i* be the smallest indices such that *a*_{i} > *b*_{i}. We can amplify the gap between *w*_{ai} and *w*_{bi} large enough to make Alice wins.

297C - Splitting the Uniqueness

An equivalent definition for *almost unique*, is an array with at least ⌊ 2*n* / 3⌋ different elements. The idea is to split *s* into three parts: In the first part, we give uniqueness to *a*. In the second part, we give uniqueness to *b*. In the third part, we give uniqueness to both.

Lets assume *s* is sorted. Since *s* is an unique array, *s*_{i} ≥ *i* for all *i* (0-based). The image below will give some intuition on how we will split it. *a* is red, *b* is blue, the length of the bar represent the magnitude of the number. In the first and second part, we do not care about the array that we are not giving uniqueness to.

We will make an example with *n* = 30.

*i* = 0... 9: assign *a*_{i} = *i* (do not care values of *b*)

*i* = 10... 19: assign *b*_{i} = *i* (do not care values of *a*)

*i* = 20... 29: assign *b*_{i} = 29 - *i*. *a* takes the remains. From *i* = 20, *a* will have strictly increasing values starting from at least 11.

For *k* = 1 there is only one coloring so we just need to check the number of "E" constraints. When *k* ≥ 2, it turns out that using only 2 colors is sufficient to satisfy the constraints.

We will call the constraints that involves cells in different row "vertical constraints", and similar for "horizontal constraints".

First of all, we color the first row such that all horizontal constraints in row 1 are satisfied. We will color the remaining rows one by one.

To color row *i*, first we color it such that all horizontal constraints in row *i* are satisfied. Then consider the vertical constraints between row *i* and row *i* - 1. Count the number of satisfied and unsatisfied vertical constraints. If there are more unsatisfied constraints than satisfied constraints, flip the coloring of row *i*. Flipping a row means turning 2211212 → 1122121, for example.

If we flip the coloring of row *i*, all horizontal constraints in row *i* are still satisfied, but for the vertical constraints between row *i* and row *i* - 1, satisfied will turn to unsatisfied, unsatisfied will turn to satisfied. Therefore, we can always satisfy at least half the vertical constraints between row *i* and row *i* - 1.

Now for each row, all horizontal constraints are satisfied (*m* - 1 each row), at least half vertical constraints with the upper row are satisfied (⌈ *m* / 2⌉). In total we satisfied *n*(*m* - 1) + (*n* - 1)⌈ *m* / 2⌉ constraints. Finally notice that when *m* ≥ *n*, we satisfied at least 3 / 4 fraction of the constraints — if it is not the case that *m* ≥ *n*, simply rotate the table.

Problem and editorial written by AEtheReal. Link to editorial.

"If n > m, we can increase every wi for sufficient amount and done, because Alice always gains more than Bob." This sounds unclear to me. I would just say that we can set all w_i to be equal to 1 and then Alice wins.

Oh, that's a simpler argument! Do you mind if I put in in the editorial?

How could I mind, go ahead.

Could you tell me how did you invent problem Div1-C? I mean where did the idea come from? :) It should be interesting! :D

Actually it does not come from real life problem. I was thinking to make a problem about splitting, and I came up this.

Oh :)

I really liked the idea :)

"(=>) Let i be the smallest one such that a_i > b_i. We can amplify the gap between w_a_i and w_b_i large enough to make Alice win."

Would it also be correct to say make w_a_i much larger than w_b_i and make w_x = w_a_i for any x > a_i, so that the rest of the terms cancel each other out?

Yes

div2-E and div1-D are awesome.

This contest is more about idea rather than code ability. I love it actually though code ability is poor.

And this place sounds unclear to me. There is not exactly one more vertical constraints than horizontal: there is

w-hmore vertical constraints. And I don't understand what are you talking about in following statement, in fact we have already satisfied 75% of constraints (of course ifh≤w).Yeah, i also think so. The number of vertical constraints is (h — 1) * w = hw — w, and the number of horizontal constraints is (w — 1) * h = hw — h. In this solution, You have already satisfied all horizonal constraints and half of the others. So, if w >= h, You have already solved the problem.

Ya, I mean, if we only consider the constraints "settled" after filling the i-th row, it may be less than 3/4. I was hopping to explain how to come up the idea of "W >= H", instead of writing "if H > W swap them" in the beginning of the solution, which may be confused "how do you come up this?"

1 ques for fish weight you said trim the last part of b and now a=b let us take the 2nd sample where a[]={2,3,5,7} and b[]={2,3,3,5} and thus a[2]>b[2] i.e ai>bi so, ans should be Yes or i am reading it in wrong way i.e using only last n samples of b?

You need to sort it in decreasing order first. So A[] = {7,5,3,2} and b[] = {5,3,3,2} and thus as you said a[1] > b[1] and so the answer is Yes.

k got it thnx :)

what is the output for div2-d if the input is :

2 2 500 1 2 500 1

if w1 = 1 and w2 = 500, then alice = 1 + 1000 = 1001, and bob = 500 + 500 = 1000, so, it should be "YES" right ?

I'm very bad at "find one solution",and the solution is very simple, like Div-C of this contest or #170 div1-B (http://www.codeforces.com/contest/277/problem/B) I can hardly invent them...

Is there any typical way of thinking? also, would someone give me some training problems like these?

I have another solution in div1-C which I prefer to share with you my solution : consider the array s sorted , now for ( 1<=i && i<=n/3 ) a[i]=0 , b[i]=s[i] ; for (n/3+1<=i && i<=2n/3) b[i]=0 , a[i]=s[i] and finally for (i>2n/3) a[i]=n-i+1 , b[i]=s[i]-(n-i+1) It can easily be shown that the sequence is almost unique.

what about this case s = {1,2,3} your answer is a = {0,2,0}; b = {1,0,3}; is this answer correct?is "a" almost unique?

I know that it is a bit later but after think a lot the problem 297D, I have the equation to solve: constraints satisfied >=0.75*(total constraints) . That is : n*(m-1)+ceil(m/2)>=0.75*(n*(m-1)+m*(n-1)) then you have m>=n That's all! (Sorry for my poor english)