minute

You could reach 2000 after this round. Set your dream high.

is it possible to achieve 1300 for me

Your rating curve is so smooth lol

m=0; c=1400; y=m*x+c;

1. Process the exponent (i) from 0 to 62 (actually 60, but it doesn't matter)

2. If the i-th bit of n is set, then find the smallest existing box that is at least 2**i, then divide it by 2 until it is 2**i, creating two other boxes in each division. Then use a box of size 2**i.

3. Then use all k boxes of size 2**i to create k/2 boxes of size 2**(i+1)

SUM = sum of all Ai. So if SUM < n answer is 0. else SUM = n + reminder. now go from big powers to small. a = 2^k. if reminder>=a so reminder-=a else if n>=a so n-=a else cnt[k-1]+=2 and start iterate on k-1.

Greedy. First for each power of 2 count how many boxes with this number we have. Then start looking at the bits of n, starting from the lowest one. If this bit is 0, we just go to the next bit, and in the meanwhile try to merge the boxes with corresponding size (2^i) to "create" boxes with a higher power of 2. If current bit is 1, we use the box of size 2^i, if possible. If it is not possible, we take the "nearest" box with bigger size and divide it until we get a box of the right size.

You just need to construct all of the bits of number 'n' . Start from constructing lowest bit .for constructing ith bit first check if it can be constructed by numbers smaller than or equal to 2^i . Else you need to divide some number 2^j , j>i and j should be closest to i , see if this is possible .if no for some bit then -1 else answer is sum of number of divisions .

of course above is just rough idea but you can go through the submission and ask if you have some doubt .

submission 70903270

1 a

I think you should check your code in the test case if the string length = 1.

Unfortunately, we had to put tight memory limits to disallow solutions with dynamic connectivity offline approach.

By the way, which approach uses $$$O(nmc)$$$ memory and cannot be modified to use $$$O(nm)$$$ memory?

You should do (1LL<<smth) instead of (1<<smth) because it overflows.

Now i got my mistake...I consider if any number x is divide by 2 then number become x/2. Then i consider only 1 frequency of x/2 But actually Now we have 2 frequemcy of x/2. I misunderstood the problem..

Which problem from this round was copied?

Got it, i'm now tring to solve ApIO18 (LOL

not slow

It's centroid decomposition with convex hull trick. As TL is 6 seconds, it should pass.

I tried too, but I had an error in my code during contest that I couldn't fix in time. However, after fixing it it gave me WA, i think I'm bad implementing the dynamic convex hull trick.

For each possible split t1 + t2 of the string t, dp[i][j] = "minimum prefix of s such that it contains non-intersecting i-th prefix of t1 and j-th prefix of t2"

First try every split of the string t and then dp[Position of current symbol in the string s][length of the first subsequence] — maximum number of symbols in the second subsequence.

(Mine is kind of weird, maybe there is a better one)

Assume the first subsequence is of fixed length N. Then dp[i][j] = after taking the first j characters of string s, it is the largest length of second subsequence, given first subsequence has length i. If dp[|s|][N]=|s|-N, then we output YES. Now repeat for all possible N. If no such answer is found, output NO.

Submission: https://mirror.codeforces.com/contest/1303/submission/70902911

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Well, variable 'n' has no certain value in your code.

for (i=0; i<n && s[i]=='0'; i++);

here variable n is not defined. Use s.size() instead.

        if(g>=b)
            cout<<n<<"\n";

i cant understands this, you can see my code

else if((temp%g)!=0)
{  
                cout<<(tp*b)+temp<<"\n";
}

You need to check if the answer is no less than n, as in the following "else" statements. (Input: 10 3 4, Output 9, Answer 10)

Imagine waking up and half of them had been hacked, just kidding :P
Good night.

If there is a cycle then there is no solution. Checking if there is someone with degree 1 is not enough.