Codeforces Edu Round 111 | Problems A to E Solutions

[@adityamonu429]

I literally wait for every editorial/explanation by you after every contest please try to make all editorials

[@ashishverma1382]

nice explanation bahut dino baad problem C samajh aaya hai

[@rishikeyyadav5717]

Great intuition for C.thnx it helped..

[@alaymehta2449]

You could have uploaded Kickstart solutions

[@tusharbk8393]

Answer of A should be ceil of sqrt of n bcoz sum of first n odd numbers is n square which will be optimal solutions

[@deepanshsharma4342]

An awesome explanation for C, thank you!

[@kishansinha9466]

Please add timestamps

[@shivamgupta917]

very clean observations Awesome!!!

[@stunnerhash]

Mannnnn i was so close to the solution to C... I figured out the logic but instead of bruteforcing i used a complex implimentation and then failed to debug it in time

[@deepanshu6751]

A was just basic observation that max sum from n numbers can be n^2 iff all n numbers are odd so just ceil value of sqrt(s) would give n. Like max S with 3 would be 9 iff 1 3 5 are in array so if we want 10 we have to have 1 more element so ceil(sqrt(10))=4 which is our answer.

[@pranavrajveer3767]

How so OTZZ

[@danilzashikhin3591]

hi

[@iampatelajeet]

Look what I did, As you can see the answer for sum n is square root n. Actually number n covers sum till n^2 from n-1^2+1 Ex: 4^2 = 16, 3^2 = 9 so for sum 10 to 16 answer will be 4 and further 5 for 17 to 25. And we know that max sum is 5000 so answer will we maximum 71. So make a loop till 71 and check which number's square is >= sum, Like every number don't have a perfect square root as sum = 24 so i = 5 is 25 which is just greater than sum and a perfect square. Hence 5 is anwer.

[@jaywagner3934]

Ceil of square root of n

[@iampatelajeet]

@@jaywagner3934 yup ✌

[@Code_Solver]

Great Bro can you remember me

[@iampatelajeet]

@@Code_Solver ohh got you. 😃✌✌