Codeforces Round 942 (Div. 1, Div. 2) Editorial

1972A - Contest Proposal

1972B - Coin Games

1967A - Permutation Counting

1967B1 - Reverse Card (Easy Version)

1967B2 - Reverse Card (Hard Version)

1967C - Fenwick Tree

1967D - Long Way to be Non-decreasing

1967E1 - Again Counting Arrays (Easy Version)

1967E2 - Again Counting Arrays (Hard Version)

Thanks A_G for discovering that E2 is possible!

1967F - Next and Prev

• 5 hours ago, # ^ | I thought the problem was cool

• 5 hours ago, # ^ | not enough persistent segment tree...

• 4 hours ago, # ^ | ← Rev. 2 →   0 it is mid night here and i had to do mAtHs

• 4 hours ago, # ^ | it had some very nice problems, definitely

• 4 hours ago, # ^ | D1 is not mine, it was accidentally invented by the tester newb_programmer while he was solving D2.

• 4 hours ago, # ^ | I liked 2F/1D. Too hard for me to solve during round but solution nicely unfolds in several steps

• 4 hours ago, # ^ | ← Rev. 2 →   +3 2F/1D is really cool. I would have been annoyed if hint 2 was the intended solution. Btw typo in editorial

#include "bits/stdc++.h"
#define int long long

signed main() {
    int tt; scanf("%lld", &tt);
    while (tt--) {
        int n; scanf("%lld", &n);
        char c[n]; scanf("%s", &c);
        std::string s(c);
        int cntU = std::count(s.begin(), s.end(), 'U');
        if (cntU % 2 == 1) printf("YES\n");
        else printf("NO\n");
    }
}

• 4 hours ago, # ^ | Inadequate char array length.

  • 4 hours ago, # ^ | ← Rev. 3 →   0 Thanks!

  • 4 hours ago, # ^ | but why? its size is n

    • 4 hours ago, # ^ | You need a \0 (which is if you convert it to int) to tell the language that a C-type string ends here. So you'll need extra space.

      • 4 hours ago, # ^ | ok thanks! i will never use char[], scanf, printf

        • 3 hours ago, # ^ | That's a bit of an overreaction, isn't it?

• 4 hours ago, # ^ | I've tried, but its making solution too much complex bcz of such input: 1 1 1 2 2 2 3 3 3

• • 3 hours ago, # ^ | can you explain please?

    • 3 hours ago, # ^ | Here curr is storing the minimum value all the elements can reach and rem stores the remaining operations after that minimum value is reached. Every element in priority queue I am comparing with previously reached value and calculating whether it possible to reach next top of priority queue value with remaining k.

• 4 hours ago, # ^ | then I fst at 1A

• 4 hours ago, # ^ | You don't need binary search if you already sorted array. Find = number full permutations (bin search or sort + prefix sum) = remaining k add (and purchase distinct ) at any end of full permutations to form additional permutations of . 258906139

  • 4 hours ago, # ^ | I used binary search on k, I sorted array just ,

• 4 hours ago, # ^ | We can also notice that must be a multiple of and iterate from and from , then it turns into .

  • 4 hours ago, # ^ | Unfortunately couldn't make the same trick work for Div2D2, do you have any idea how?

    • 3 hours ago, # ^ | ← Rev. 2 →   0 My solution for D2 uses same trick 258920365. Though it is O(n log^3(n))

      • 3 hours ago, # ^ | Cool that it fits in time

        • 2 hours ago, # ^ | Well, i am sure that it is less than O( nlog^3 n), but i don't know what it is exactly.

• 2 hours ago, # ^ | your complexity is :) max answer is

  • 2 hours ago, # ^ | Thanks. Is this explaination correct? "For most cases, gcd(u,i-u)=1, so we can assume the enumeration quantities approximate to the answer."

• 4 hours ago, # ^ | mid=1e17, and k-=mid*n.

  • 4 hours ago, # ^ | Thanks i got it ,k can get too much negative to handle more subtraction

  • 4 hours ago, # ^ | but how can i fix this?

    • 4 hours ago, # ^ | break when k<0

• 3 hours ago, # ^ | Take a look at the codeforces catalog. There people much smarter than me have posted the exact thing you are looking for. From resources to guides on how to improve and tutorials on specific topics and even stuff on mindset and attitude. https://codeforces.com/catalog

• 78 minutes ago, # ^ | Here is an optimal solution for that test case Buy 3 cards of the first and third ones Buy one card for the fourth one Buy two cards for the fifth one then you can arrange them as follows and get 27 as the maximum score:

• 15 minutes ago, # ^ | Why didn't you mention in the comment that it's not in English? To gain some unintended view?

• 6 minutes ago, # ^ | Inside the binary search, you can break when tmp < 0 (clearly). Turns out, you "must" break otherwise some negative integer overflow will occur and unintentionally make tmp positive even though it should DEFINITELY be negative once tmp < 0 occurred. I got AC on your submission with this change. 258943287

  • 4 minutes ago, # ^ | yeah i just found the problem with my code

• 5 minutes ago, # ^ | NVM it was actually signed integer overflow! :(