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Codeforces Round #341 (Div. 2) Editorial
By minimario, history, 9 years ago, In English

621A - Мокрая Акула и чётность

First, if the sum of all the numbers is already even, then we do nothing. Otherwise, we remove the smallest odd number. Since, if the sum is odd, we need to remove a number with the same parity to make the sum even. Notice it's always bad to remove more odd numbers, and it does nothing to remove even numbers.

621B - Мокрая Акула и слоны

Let's start with two bishops (x1, y1) and (x2, y2). Notice that if (x1, y1) attacks (x2, y2), either x1 + y1 == x2 + y2 OR x1 — y1 == x2 — y2. So, for each bishop (x, y), we will store x + y in one map and x — y in another map.

621C - Мокрая Акула и цветы

Let f(x) be the probability that the product of the number of flowers of sharks x and is divisible by p.

We want the expected value of the number of pairs of neighbouring sharks whose flower numbers are divisible by p. From linearity of expectation, this is equal to the probabilities that each pair multiplies to a number divisible by p, or f(0) + f(1) + ... + f(n). (Don't forget about the wrap-around at n)

Now, for each pair of neighbouring sharks, we need to figure out the probability that their product is divisible by p. Consider an interval [li, ri]. How many numbers in this interval are divisible by p? Well, it is easier if we break the interval [li, ri] up into [1, ri] - [1, li - 1]. Since 1, 2, ..., x contains numbers divisible by p, the interval [li, ri] contains numbers divisible by p.

Now, consider two numbers and , with . Let ai be the number of integers divisible by p in the interval [li, ri], and define aj similarly. Now what's the probability that fi·fj is divisible by p? We can count the opposite: the probability that fi·fj is not divisible by p. Since p is a prime, this means neither fi nor fj is divisible by p. The number of integers in [li, ri] not divisible by p is ri - li + 1 - ai. Similar for j. Therefore, the probability fi·fj is not divisible by p is given by . Therefore, the probability it is can be given by . Now, just sum over this for all i.

621D - Крыса Квеш и сыр

The tricky Rat Kwesh has finally made an appearance; it is time to prepare for some tricks. But truly, we didn't expect it to be so hard for competitors though. Especially the part about taking log of a negative number.

We need a way to deal with xyz and xyz. We cannot directly compare them, 200200200 is way too big. So what we do? Take log! is an increasing function on positive numbers (we can see this by taking , then , which is positive when we are dealing with positive numbers). So if , then x ≥ y.

When we take log, But yz can still be 200200, which is still far too big. So now what can we do? Another log! But is it legal? When x = 0.1 for example, , so we cannot take another log. When can we take another log, however? We need to be a positive number. yz will always be positive, so all we need is for to be positive. This happens when x > 1. So if x, y, z > 1, everything will be ok.

There is another good observation to make. If one of x, y, z is greater than 1, then we can always achieve some expression (out of those 12) whose value is greater than 1. But if x < 1, then xa will never be greater than 1. So if at least one of x, y, z is greater than 1, then we can discard those bases that are less than or equal to 1. In this case, . Remember that , so . Similarly, .

The last case is when x ≤ 1, y ≤ 1, z ≤ 1. Then, notice that for example, . But the denominator of this fraction is something we recognize, because 10 / 3 > 1. So if all x, y, z < 1, then it is the same as the original problem, except we are looking for the minimum this time.

621E - Мокрая Акула и блоки

First, let us build an X by X matrix. We will be applying matrix exponentiation on this matrix. For each modulo value T from 0 to X — 1, and each value in the array with index i between 1 and n, we will do: matrix[T][(10 * T + arr[i]) % X]++. This is because, notice that for each block we allow one more way to go between a modulo value T, and (10 * T + arr[i]) % X. We are multiplying T by 10 because we are "left-shifting" the number in a way (i.e. changing 123 -> 1230), and then adding arr[i] to it. Notice that this basically simulates the concatenation that Wet Shark is conducting, without need of a brute force dp approach.

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9 years ago, # |
← Rev. 2   Vote: I like it +16 Vote: I do not like it

How foolish i am, i thought of the Problem C's Independence.at the end of the contest i finally figure out that if ppl_2 have a number can be divisible by p,we dont need to care about what number ppl_1 and ppl_3 take:(

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9 years ago, # |
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D is more difficult than E...

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    9 years ago, # ^ |
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    That's ridiculously common...

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    9 years ago, # ^ |
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    Probably it's just because DP/Matrix exp. are well-known and heavily trained topics, whereas D is somewhat non standard. I solved D in 12 minutes and E in more than 1 hour... Though I used python's decimal for D which is a bit hacky.

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9 years ago, # |
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in Problem C why did you say x*(x+1) mod n divisible by p?

isin't x(x+1) divisible by p?

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9 years ago, # |
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Problem E can be solved without matrix exponentiation. Denote by DP[b] the solution array (i.e. DP[b][k] is the actual solution, the others are needed for the computation). Now, if we compute DP[2 ^ i], for all i < 31, we can easily get DP[b] by making use of the binary representation of b. The whole solution hinges on a function Unite(a, b), which computes DP[a + b] from DP[a] and DP[b] in time O(x ^ 2).

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    9 years ago, # ^ |
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    same solution

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    9 years ago, # ^ |
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    There my solution (runtime somewhere, but guess idea is correct)

    denote D1[d][k] count d-digit numbers with k-reminder. Easy to calcuate by:

    D1[d][(10 * j1 + j2)] += cnt[j1] * D1[d — 1][j2]; 0 <= ji < x;

    Now make blocks 1000, 1000, 1000,.... b % 1000 digits (b / 1000 + !!(b % 1000) blocks);

    denote D2[d][k] count d-block numbers with k reminder. Calculate absolutely analogically.

    Then ans is D2[blockCount — 1][k]. Complexity is O(1000*x^2).

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      9 years ago, # ^ |
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      If i'm thinking correct, complexity would be O(10^6 * x^2). You'll make D1 in O(1000*x^2) but then there can be 10^6 blocks of 10^3 length . If you again divide your 10^6 blocks into 10^3 "big" blocks , then it should work.

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        9 years ago, # ^ |
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        Thanks for correcting, I'm "the genius of Math":D

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    9 years ago, # ^ |
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    Can you elaborate slightly on your approach?

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9 years ago, # |
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What exactly are we doing with matrix exponentiation in problem E? Can somone please explain. I am not able to understand the editorial solution!

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    9 years ago, # ^ |
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    Optimize dp.

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    9 years ago, # ^ |
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    To simplify the explanation, it is assumed that all computations are done modulo 1, 000, 000, 007.

    First, note that the n given digits are not important -- what is important is the occurrences of each digit. Let occ[d] be the occurrences of digit d.

    We will try solving this problem using DP. Let dp[i][j] =  the number of ways we can pick i digits such that the final modulo is j.

    The base case is obviously dp[0][0] = 1. For i > 0, the recurrence can be given as:

    dp[i][j] = sum{occ[d] × dp[i - 1][a]}, for all 1 ≤ d ≤ 9, and for each d, a is an integer 0 ≤ a < X where (a × 10 + d)%X = j.

    (Intuitively, given a number consisting i - 1 digits that is a modulo x, we can form a number consisting of i digits that is a × 10 + d modulo X = j, by appending the digit d.)

    The final answer would be dp[B][K]. Unfortunately, this DP will get TLE as B can be up to 109. To speed it up, we will use matrix exponentiation trick on the recurrence.

    Suppose we have a vector F0 consisting of X elements, where

    .

    Exercise 1: Compute the value of the elements F0!

    Now, construct a X × X matrix T, in such a way that

    F1 = TF0,

    where

    .

    Exercise 2: Compute the value of the elements of T!

    In a similar way, we can compute F2 = TF1 = T2F0. Now, we want to compute

    FB = TBF0,

    which is,

    .

    The answer will be... FB[K], which is dp[B][K]!

    TB can be computed in using exponentiation-by-squaring, which will fit the time limit.

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9 years ago, # |
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D can be solved with complex numbers, to avoid having to treat cases where some numbers are smaller than one, differently. Note that the real part of log(log(x)) is growing if log(log(x)) is real, and decreasing if it has an imaginary part. So you can just write a comparison function to find the maximum.

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    9 years ago, # ^ |
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    Awesome, now we can easily handle such cases that log(x) is less than zero. Thanks for your idea.

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    9 years ago, # ^ |
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    great idea!

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    9 years ago, # ^ |
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    Can you explain how to compare between complex numbers?

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      9 years ago, # ^ |
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      If x > 1, then log(log(x)) is an increasing function, and if x < 1, then real(log(log(x))) is a decreasing function, because taking a logarithm of a negative number results in something like this: log( - x) = log( - 1·x) = log( - 1) + logx = iπ + log(x). (Assuming log(x) is done in base e) Therefore, to compare two numbers by their loglog, you can do something like this:

      bool compare (complex x, complex y) {
  if (imag(x) == 0 and imag(y) == 0) 
    return real(x) > real(y);
  else if (imag(x) != 0 and imag(y) == 0) 
    return false;
  else if (imag(x) == 0 and imag(y) != 0) 
    return true;
  else if (imag(x) != 0 and imag(y) != 0) 
    return real(x) < real(y);
}
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    9 years ago, # ^ |
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    "long double" is still required,because the result of y^z^x will exceed the range of "double" on (test 9: 185.9 9.6 163.4 ) when using "complex"

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      9 years ago, # ^ |
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      You shouldn't ever actually calculate the value of y^z^x, since log(log(yzx)) = log(zxlog(y)) = xlog(z) + log(log(y)). That's the whole point of using log log, you can compare those gigantic numbers without actually calculating them.

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        9 years ago, # ^ |
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        I made a mistake. :< I forgot to modify the calculation function.I just calculated log(y^z^x).

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9 years ago, # |
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Is there a good matrix library in C++? Or you simply write an algo for matrix binpow each time you need it?

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9 years ago, # |
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But yz can still be 200200, which is still far too big.

Apparently, one of the passed solution use that method.

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    9 years ago, # ^ |
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    Interesting. How it works? In C# we will just get Infinity value in such case, and Infinity is equals to another Infinity, so wrong result on case #9.

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      9 years ago, # ^ |
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      Codeforces compiler suggest that maximum range of long double is 1.18973 × 104932,  which is greater than 200200 ≈ 1.6 × 10460

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        9 years ago, # ^ |
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        Ouch, nice. In C# max for double is just 10^308.

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      9 years ago, # ^ |
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      C# also has BigInteger type. Wouldn't be suitable for this particular problem, but very useful for others.

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9 years ago, # |
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Can somebody help me with this 15713106 submission for C? I cannot understand why it failed.

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9 years ago, # |
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E can be done in too- let vj be the array of size x such that vj[i] is the number of ways of getting a number that is i mod(x) after j steps; v2j and v2j + 1 can be computed from vj in . The crucial fact here is that if the value of the first p blocks is i mod(x) and that of the last q blocks is j mod(x), then the whole number will be i + 10q·j mod(x).

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    9 years ago, # ^ |
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    You mean the whole number is (i * 10 ^ q + j mod x) since i comes first .

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9 years ago, # |
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can anyone explain C in easier way

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    9 years ago, # ^ |
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    The intuition is that if you can find out the probability of each person, then you can solve the whole problem. This is because the whole sum of money is the expectation of each cases(1-2, 2-3, etc..). The expectation of each pair is 1 — (1-p_i)(1-p_j) because the pair would not get paid only when both don't satisfy the condition. This is same for all the pairs, so now we just need to find out the probability(expectation) of each person. This is explained in the editorials

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      9 years ago, # ^ |
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      thank you so much but i want to ask if you can tell me about a good resource to learn about this kind of problems

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        9 years ago, # ^ |
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        I'm not sure if your "these kind of questions" meant algorithm questions is general, or this particular question, but assuming the latter I recommend studying basic discrete mathematics. This set especially had a lot of math questions, and understanding of discrete mathematics will make you have a better intuition of solving problems. The more you learn, the more you'll see.

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9 years ago, # |
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Why this approach is wrong for B

iterate each diagonal of the grid and count the number of bishop x in that diagonal .

if the number of bishop is > 1 than we add x(x+1)/2 to the answer .

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    9 years ago, # ^ |
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    i get accepted with the same idea you may made a mistake in the code http://codeforces.me/contest/621/submission/15705229

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      9 years ago, # ^ |
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      This is my submission

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        9 years ago, # ^ |
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        It has to be x(x-1)/2 rather than x(x+1)/2. A.Magdy7's solution works because he decreases x by 1 before calculating with the x(x+1)/2 formula.

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          9 years ago, # ^ |
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          yeah you are right , silly mistake but even when i changed x by x-1 its still give WA .

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            9 years ago, # ^ |
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            Nope.

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            9 years ago, # ^ |
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            Consider there are x bishops in a diagonal. The first bishop has x-1 bishops to pair with, the second bishop has x-2 bishops to pair with, and so on. We can notice the number of pairs is (x-1)+(x-2)+...+1 which can also be expressed as (x-1+1)(x-1)/2 = x(x-1)/2.

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          9 years ago, # ^ |
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          because you don't search in all the diagonals you always search above the main diagonal in both direction and don't search under it

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9 years ago, # |
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My solution of C : First we see how to find the expected money given to any two neighbouring sharks :

For shark1 -
int p1 = count of numbers between l and r which are divisible by p
int n1 = count of numbers between l and r

Now, p1 = r/p - (l-1)/p
     n1 = r-l+1  
Similarly define and calculate p2 and n2
Now, money is given to both the sharks only when s1*s2 is divisible by p . 
This is possible in following cases :
 i. s1 & s2 both are divisible by p . Count of such numbers is p1*p2 
 ii. Only s1 is divisible by p . Count of such numbers is p1*(n2-p2)
 iii. Only s2 is divisible by p . Count of such numbers is p2*(n1-p1)

Since, 
 i. Count of all the different s1*s2 that can be obtained is n1*n2
 ii.For every number s1*s2 divisible by p each shark get 1000 dollars (2000 total)
Hence, expected money that wet shark gives is
 2000( p1*p2 + p1*(n2-p2) + p2*(n1-p1) )/(n1*n2) = x (say)

We calculate this x for the n+1 pairs and add them . This is our final answer . 
Hope, it helps.

Here, is my code which i couldn't submit because of internet issues : Code

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    9 years ago, # ^ |
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    Can you explain something about linearity of expectation?

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9 years ago, # |
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So Java's sorting does some sketchy things, but only some of the time...lol

http://codeforces.me/contest/621/submission/15716716 AC

vs

http://codeforces.me/contest/621/submission/15716729 TLE

with the only difference switching 

Arrays.sort(a);

with the unrelated

if(tot%2==0){
	System.out.println(tot);
	return;
}

    `

At least my rating is historic now!

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    9 years ago, # ^ |
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    I'm pretty sure this is because Arrays.sort calls quick sort for primitive data types and merge sort for objects. So there must have been an anti quick sort test case, in which quick sort is called for one solution but not the other.

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9 years ago, # |
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Why can't I open problem D? it says : Can't read or parse problem descriptor. Is any one else getting this error?

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9 years ago, # |
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Why my code although passing test-case 9 of 100000 numbers but getting TLE on test-case 15 of 500 numbers http://codeforces.me/problemset/problem/621/A http://ideone.com/OWI8Ba

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    9 years ago, # ^ |
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    i=n-1; in each iteration, cycle infinity, for example, in this test: 3 2 3 4

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9 years ago, # |
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"But y^z can still be 200^200, which is still far too big" Can anybody help to find a test that breaks my solution 15713937 ?

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9 years ago, # |
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For D: Python's decimal for the win :) one logarithm is enough

15703568

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9 years ago, # |
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Could anyone help to understand what is wrong here with Problem D?

  • submission with EPS = 1e-14 gives wrong answer on test #134: 200.0 200.0 200.0: 15719829,

  • the same code with EPS = 1e-13 gives AC: 15719618.

I've logged the difference between two values of m * log(m) + log(log(m)) for m = 200.0 in first case and it appears to be ~5e-14, so this is the reason. But when I'm trying to reproduce this problem manually (see main at 15719829), it is giving zero difference. On my mac with clang 6.0 compiler case 200.0 200.0 200.0 gives proper answer: x^y^z.

It is OK that taking of two exactly the same arithmetic expressions with "equals" double numbers can give a different answer?

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    9 years ago, # ^ |
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    15735486 I change left + EPS < right to left — right < 0. Also you can use long double instead double.

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      9 years ago, # ^ |
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      Thank you! But another interesting thing is that solution with 1e-14 passed when I changed compiler on the MS Visual Studio 2010 on CF: 15735359.

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9 years ago, # |
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Calculating C the probability itself (and not the 1-p) gives WA because of error approximation.

I calculated the probability of every single position getting a multiple of the prime (multiplied by 4000) and, using inclusion-exclusion principle, removed the the cases where two consecutives where primes (multiplied by 2000).

If someone could explain me why this is incorrect or if the jury had a tighter error, I'd be very grateful.

I don't know how to post links, but the submission was: 15706251

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9 years ago, # |
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In test 69 of problem D x=0.2, y=0.7, z=0.6. My program output (y^z)^x on this test. But there is expected (y^x)^z. But (0.7^0.6)^0.2>(0.7^0.2)^0.6! I have checked this on the calculator. (0.7^0.6)^0.2-(0.7^0.2)^0.6=1.9535924907431949702433539286085e-38.

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    9 years ago, # ^ |
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    In fact, (0.7^0.6)^0.2=(0.7^0.2)^0.6 since (a^b)^c=a^(b*c). Thus you do the one with the smaller index.

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9 years ago, # |
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I bet D is simpler, so left no time for E.....

Problem D can be coped with in this way: notice that log log x^y^z = z*logy + log log x, log log (x^y)^z = yz log log x; so the only problem to discuss is log log x. if log x is positive, that is easy, cuz f(x) = log log x is a strictly increasing function, so the larger the better. if log x is negative, we take log -log x for the log log x part. so the function became strictly decreasing, and since two cases use different functions, they can not be compared. And then the only problem is the order.

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9 years ago, # |
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In problem C : Aren't f(i) and f(i+1) dependent? Is linearity valid even if they are dependent?

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    9 years ago, # ^ |
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    Yes. That's called "linearity of expectation".

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      9 years ago, # ^ |
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      I am a bit confused. You mean to say that f(i) and f(i+1) are dependent?

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        9 years ago, # ^ |
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        They are dependent, but their expected sum is not -- E[f(i) + f(i+1)] = E[f(i)] + E[f(i+1)], so you can compute the values independently.

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          9 years ago, # ^ |
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          Thanks for clearing my head. :)

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9 years ago, # |
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I failed main test C, just because I haven't changed the answer into double, and printed 6 decimal digits behind the point :(

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9 years ago, # |
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How does this solution get accepted ?

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    9 years ago, # ^ |
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    Apparently, long double covers a large portion of the numbers if thats what you are wondering about. Even larger than 200200

    Check this snippet Credit: S.Saqib

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9 years ago, # |
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Hey, everyone?

I am still unable to fully understand the explanation for problem 3. I think that there must be some sort of relation between pair (i,i+1) and pair (i,i-1) (and/or pair (i+1,i+2)) as if s[i]*s[i+1] is divisible by p, then s[i+1]*s[i+2] and/or s[i-1]*s[i] will be divisible by p. How can we just sum them up independently like that?

Could someone please elaborate on this case for me? Thank you.

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    9 years ago, # ^ |
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    You can iterate through the array, calculating expectation for each element, thus you will get the relation you're looking for. You need to handle 2 cases: when the selected "shark" gets a number divisible by p, and when it does not, but one of it's 2 neighbours does. You can check out my code if needed

    Oh, i see your point, you're asking why we can handle each pair separately. See, E = x[1] * p[1] + x[2] * p[2] + .., where E — the expected value, p[i] — probability with which x[i] will be added to the answer. The thing is, the mathematical expectation can be easily divided into independent terms, because in fact, expectation is a linear polynomial, and it's distributive property allows us to calculate expectation for sub-problems, summing up the resulting values. I mean, if x[i] = x[i]' + x[i]'' -> p[i] * x[i] = p[i] * x[i]' + p[i] * x[i]'' Therefore, we can divide the sum for the whole circle into sum of pairs' answers

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      9 years ago, # ^ |
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      Oh thank you so much, the trouble seems to come from my limited knowledge of mathematics. Your comment helps me figure out that it is linearity of expectation. I will do some googling on it.

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      9 years ago, # ^ |
      ← Rev. 38   Vote: I like it +5 Vote: I do not like it

      "can be EASILY divided into independent terms, because in fact, expectation is a linear polynomial, and it's distributive"

      I'd like to give one more level of intuition upon that, cause I saw exactly that in the wikipedia and for some reason it haven't satisfied me :)

      But, before telling the whole story and laying out all of my thoughts, I'd like to test my metaphors. My metaphors work for me, but they might not work for you. If it seems appealing to you, I will continue to end the story. If not... well, at least I've tested my hypothesis :)

      Let's try to establish the similarity of the problem to the sequence of 8 bits like that: 01110010.
      If we stretch our imagination we can see the relationship is as follows: when the pair of sharks grows good amount of flowers the Wet Shark gives away 1, when they grow the bad amount of flowers the Wet Shark gives away 0.
      The total amount of money that the Wet Shark gives away in the case 01110010 is 0 + 1 + 1 + 1 + 0 + 0 + 1 + 0 =  4. The maximum amount of money that the Wet Shark can possibly give away is 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =  8  * 1 =  8.

      Imagine that the Wet Shark should prepare it's money beforehand in the bags for money. How many bags should the Wet Shark prepare? These are all the possible sums = {0, 1, 2, 3, 4, 5, 6, 7, 8}.

      Now there are a lot of bags the Wet Shark prepared, but, to the time the flowers grow up, it would be nice to have the bag he needs to give away be right before him, so that the Wet Shark doesn't need to look for the right bag among all of those in the pile.
      So, what is the bag that the Wet Shark should expect to give away? To know this, we have to know how many of the possible events force the Wet Shark to give away the bag labeled with some number X.
      Let's go through all these bags one by one and see how many ways there are to arrive at each amount.
      For bag labeled X = 0 there is only one way to arrive at it, namely 00000000.
      For bag labeled X = 1 there are  = 8 ways to get it:
      10000000,
      01000000,
      00100000,
      00010000,
      00001000,
      00000100,
      00000010,
      00000001.
      For bag labeled X = 2 there are  = 28 ways to get it (without details :) )
      For bag labeled X = 3 there are  = 56 ways to get it.
      For bag labeled X = 4 there are  = 70 ways to get it.
      For bag labeled X = 5 there are  = 56 ways to get it.
      For bag labeled X = 6 there are  = 28 ways to get it.
      For bag labeled X = 7 there are  = 8 ways to get it.
      For bag labeled X = 8 there is again only one way to get that result: 11111111.

      So, now we can make up the function P which has as an input the label on the bag X = {0, 1, 2, 3, 4, 5, 6, 7, 8} and returns the number of ways to force the Wet Shark to give that bag of money.
      P(x) =  :
      P(X = 0) = 1
      P(X = 1) = 8
      P(X = 2) = 28
      P(X = 3) = 56
      P(X = 4) = 70
      P(X = 5) = 56
      P(X = 6) = 28
      P(X = 7) = 8
      P(X = 8) = 1

      The function P could be extended to all natural numbers by returning 0 to all of the inputs except for the inputs in the range 0..8, e.g.:
      P(X = 115) = 0 <- There are zero ways to force the Wet Shark pay 115 dollars.

      By the way, a good read about the linearity of the expected value: an intuitive explanation for the linearity of expectation.

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

It's a dumb mistake.

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