Of course not, the hacking phase starts at 0:35 in Hong Kong, but I'm a lazy guy who wakes up at 12:00 everyday XD.

do you now one time in edu contest i submit a code that shoud get TLE but after hours nobody hacked me
so i said to one of my freinds that my code is wrong after that he hacked me :\ the cool thing is that i submit that code with another handle and it got AC ;\

Honestly, I understood the meaning of your joke only a couple of minutes after downvoting your comment XD XD XD

@Java

I agree! A was especially interesting. ;)

How did u solve D?(I've got WA on 32nd testcase and I am mad).

ofc, you can't submit anymore

Since the solutions are visible discussions should not be a problem.

Yes.

Yes

Greedy. First you have to calculate appearance of each $$$a_{i}$$$ and multiply it with $$$a_{i}$$$. Then sort both of arrays and reverse $$$a$$$ (or $$$b$$$). Answer is $$$\sum\limits_{1 \le i \le n} a_{i}*b_{i}$$$

Think in fixed values, the array $$$a$$$ and how many times the number $$$a[i]$$$ (0 index base) appears in the sum expression, so you can create an new array $$$staticval$$$. Note that $$$staticval[i]=a[i]*(n-i)*(i+1)$$$ for each $$$0 \leq i \leq n-1 $$$, then you arrangue the $$$b[i]$$$ values as you need (sorting $$$b$$$ and $$$staticval$$$), note that $$$staticval$$$ array have values less than $$$10^{18}$$$, so you always can obtain the answer.

Assume that you have the optimal array as C. So we have to find the minimum of \sum_{i = 1}^n(i)*(n-i+1)*a_i*c_i (This is because the i'th element will occur when l can be anything from 1 to i and r can be anything from i to n so i'th element will occur in i*(n-1+i) number of times). So we can rearrange b_i to form C. So to get the total minimum sort the modified A array where a_i = a_i * i * (n-i+1). and array B. And then for the minimum multiply a_i & b_{i-1}. Here is the code for this.

for example: 1 2 2 8

multiply a[i] with its query frequency, sort the two arrays in reverse order of each others then calculate sum of a[i]*b[i]

-i constructed mul[i] array as the freguency of each element times a[i], mul[i]=a[i]*(n-i)*(i+1) {my i starts from 0}; -then sorted mul[], and b[] -reverse b[] -k=(k+b[i]*mul[i])%M

Assume that you have the optimal array as C. So we have to find the minimum of \sum_{i = 1}^n(i)*(n-i+1)*a_i*c_i. So we can rearrange b_i to form C. So to get the total minimum sort the modified A array where a_i = a_i * i * (n-i+1). and array B. And then for the minimum multiply a_i & b_{i-1}. Here is the code for this.

a[i]*b[i] can overflow,can do (a[i]%mod)*(b[i]%mod)

you don't mod after you multiply with the frequency

You search for bugs in codes of other people

http://codeforces.com/help#q3

Start in the first position anf, greedly, start checkin if s[i]==s[i+1], if its equal, erase s[i] and chande the odd positions to even positions, this is the answer if the string is even, otherwise, erase the last position

for every odd index, if(str[i]==str[i+1]) just remove str[i]

after that, if str.size() is odd then remove last index

I have encountered it too. I didn't try Collections.sort(). I'm getting mad. Just can't understand why Arrays.sort() gives time limit exceeded verdict.

Arrays.sort() uses the quicksort algorithm, whose worst-case running time is O(n*n). The other day one of my solutions for a Div.3 problem was hacked because someone created a counter test on Java's sort method, from this time on I coded a custom mergesort function with guaranteed O(n log n) performance.

Read this: How to sort arrays in Java and avoid TLE

In my job I am practicing java and I am facing...

your E is wrong you shouldn't mod after multiplying with frequency

You cant % array A before sorting, it gave me Wrong Answer ...

In problem F, on i th day can i order more than 1 types of micro-instruction?

For problem D can someone explain the formula for the number of divisors? p1^e1 * p2^e2 * ...

I have a small doubt in this problem F1,F2. Can we buy more than one types of microtransactions in a single day? e.g. 1 of type 1, and 1 of type 2 etc. Or its just that we can buy any number of any SINGLE microtransaction each day?

From your code it seems that we can actually buy multiple types. But this was not that clear from the problem statement.

In F the question doesn't ask us to keep the money spent minimum, so why don't we buy all type on the first day?

you have written,

if(n==1) cout << v[0]*v[0] << '\n';

which is not always true.

Example:

1 100000

Read the problem.

It's because you are not allowed to permute array a, only b. There are also other errors with your solution. You are not adding to the final answer correctly. Check this.

do exaclty the same but not sorting A before you do the algorithm above

rupav Sort A after multiplying with the coefficient (i*(n-i+1)) i.e. Sort the Array ModifiedA where ModifiedA[i] = A[i]*i*(n-i+1).

See this solution.

No the pretests were very weak for A. And to prove my point I hacked you. (Sorry)

Yes, lowerbound(set.begin, set.end, value) will take O(n). Instead you should use set.lower_bound(value)

Check this test case: Input: 1 3 3 Output: 3 Your output: 2 Set erases not unique elements

ORZ ORZ

11 5 2 11010000101

I feel like Vovuh should give you a rebate for those lost points man.

Hacking is a part of contest. If pretests are made very strong then what is the point of hacking?

I think so

If you reorder the elements of b in this way:

9 2 3 9 7

You'll have the next sums:

$$$ f(1,1) = 9, f(1,2) = 25, f(1,3) = 46, f(1,4) = 64, f(1,5) = 92 $$$

$$$ f(2,2) = 16, f(2,3) = 37, f(2,4) = 55, f(2,5) = 83 $$$

$$$ f(3,3) = 21, f(3,4) = 39, f(3,5) = 67 $$$

$$$ f(4,4) = 18, f(4,5) = 46 $$$

$$$ f(5,5) = 28 $$$

Then the total sum will be:

646

P.D. I hope this has helped you better understand the test case.

Just wait, it should be updated soon

For 6, the almost divisors will be 2, 3. Whereas 9 will only has 3 as almost divisors.

does it matter if the contests is unrated

I don't know nobody said it's unrated

System still calculating something, my standing first dropped from 500 to 1700, now it's changing between 1500-1700

No

After

if divisors[i] * divisors[~i] != guess: print(-1) break

You should continue to see other test's instead you just break I think

I had similar solution and it got accepted with Pypy3, cases like 2 999389 999917 too much for Python. Here I just wrote an alternative solution: link

I think it's already done

Yes Just who under 1600

using cf-predictor you can know your rating change will be +139 if any solution don't get wa on system test.

No, I think it's not as written in the rules...but it's very late this time..hope it will be updated soon.

still some system test is going on