Would love to know what vovuh feels after reading such cringey comments/memes.

what if i asked who is this vovuh and why is he referred here ..so many times

Go to this Link or wait for ending ICPC Challenge 2020 then can easily register from contest option..

Its because you have not participated in any contests and had less than 10 submissions....How will you differentiate then???

I cannot believe what you said.

• Nobody:
• Somebody: -- "Nobody":

.

Mine will probably start bugging me on purpose if I say that.

Try some new idea :))

So, was it true?

Conspiracy theory :p

Stupid comparison

What's your feeling today can solved not only D also E1 .

"It wasn't important"

Checkout this blog post

Div-3/educational round Hack phase :
suppose think. A test can fail a problem .But that was not in the pretest system case.So that problem showed accepted..Then anyone can hack that problem by that case in the 12 hours hacking phase.And the defender no of solved problem-=1. But hacker did not get any point.

as I know you are a candidate (maybe I am wrong)

I think you don't need Div. 4 contests :|

Conspiracy theory

In my opinion no. A-E1 obvious, E2, F — impossible.

D problem was giving TLE when i used unordered_map and when i changed it to map it got accepted. You just have to calculate the minimum number which we have to add to the numbers of the array after which they are divisible by k.

This can be calculated by taking mod.

for example the numbers are 5,1,3,4 and value of k = 3

so for the first element we need to add 1 to get the divisible number

this number can be calculated by taking mod of a[i]-k & k.

-> (k + (a[i]-k)%k)%k;

now similarly for other numbers also we'll calculate in the same way so -> 1,2,0,2 is the required array.

Now Notice that at a time we can select only one element from the above calculated array. It means that when we see 2 or more elements of same type we can't give the same x to them.

It means we have to calculate the next greater number which must be added so that the given number is divisible by k.

the next number can be calculated by multiplying k with the number of times this number has previously occurred, which means we have to maintain the hash table for it.

Your code here...
        long long int n,k;cin>>n>>k;
        long long int a[n],maxm=-1;
        for(int i=0;i<n;i=i+1) scanf("%lld",&a[i]);
        
        map<long long int,long long int>mp;
        vector<long long int>rem;
        
        for(int i=0;i<n;i=i+1)
        {
           long long int val = mod(k-a[i],k);  // taking mod and storing it.
            if(val)rem.push_back(val);
        }
        for(int i=0;i<rem.size();i=i+1)
        {

            long long int val = k*mp[rem[i]] + rem[i];
            mp[rem[i]]++;
            maxm = max(val,maxm);
            
        }
        maxm++;
        printf("%lld\n",maxm);

My approach:

for all the array elements calculate the value required to make it a number divisible by k and store it in map and keep incrementing its value if it repeats.Then, find the max "value" in the key-value pair and ans= key+(value-1)*k+1. If two keys have the max value, take the max key.

eg: second example, 5 repeats 3 times. so ans= 5+(3-1)*6+1=18.

https://codeforces.com/contest/1374/submission/85372487

Isn't this linear ... why it's giving TLE?

I think these would easily be div2 E/F.

raj1307

What Exactly you find amazing in them??

nice, 2500 problem for div3 contestants

I tried that but I was failing at test 12. Well if this worked for you then I definitely made some really stupid mistake.

Try this test case

1
5 38
15 29 7 15 7 

Edit: answer should be 70. Your code gives 108.

Just simulate. If it does not end in correct ordering, then do one triple swap including two same elements from a[]. The simulate again. If still not correct ordering it is impossible.

I don't think your code is too long xD https://codeforces.com/contest/1373/submission/85026680

We don't specify the exact time limit for compilation, but your solution can't be compiled on my laptop in 30 seconds or even in 1 minute. Probably, it is a bug of GCC or how your solution is written. Try to fix it and submit again.

what if they only read from both books. what would be your answer, you will print INT_MAX which is wrong

what if both is empty?

Even I did the same priority_queue thing. But got stuck at test case 5.

I failed at tc 4 too with priority_queue too because i forgot to convert it into the ascending one.

If you really think so, then try this

I think instead of using 3 pointers, you can just iterate over how many books are you taking which Alice and Bob both like. Then it is obvious that you have to take k — (both likes) books from 1 0 and 0 1. Just precalculate the prefix sums.

You could have implemented it directly using priority queue, one for each of the possible cases.

Then greedily go through the input :)

Hi! I think the approach for E1 was fairly straighforward. (maybe)

I stored time for (0,1) , (1,0) and (1,1) in different arrays and sorted them by non decreasing order of time.
Also we don't need to store (0,0) (why?)

Say we can take x amount of (1,1) then we know we have to take k-x from (0,1) and (1,0).

Let's iterate over all possible case of x and the minimum possible time is answer.

Here's my submission : 85342055

Thanks!

You overkilled it.

https://codeforces.com/contest/1374/submission/85352661

may be your while(1) loop run for long time(>2000ms)

Try using map instead of unordered_map. Because unordered map has a worst case complexity of O(n), your code has a worst case complexity of O(n^2).

Can you explain me C I feel so stupid :(

use map instead of unordered_map

I was thinking that unordered_map is faster than map? Can anyone make me clear?