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Topic 175 Posts

codeforces

Solutions for Codeforce。Include Div2 and Div3. Include Div1 ABC(<2500) also.
Newer Posts Page 21 of 22 Older Posts

Codeforces Round #698 Div2 Summary - Construction Problems

By xloypaypa in codeforces on 03 Apr 2021

All construction problems. The construction problems test the thinking ability and IQ the most. But the problem is that I'm not good at both of it. I'm 蒟蒻.…

Codeforces Round #698 Nezzar and Nice Beatmap - Computational geometry

By xloypaypa in codeforces on 03 Apr 2021

Solution Let's consider three points first. three point can build a triangle. So at most 1 angle equal or more than $90^\circ$.…

Codeforces Round #698 Nezzar and Binary String - Segment tree

By xloypaypa in codeforces on 03 Apr 2021

Of course a segment tree problem. Let's start from $f$. Then actually, the state of last day is fixed. Also fixed for the day before last day. The only thing we need to know is how many zeros between $l$ to $r$. Then we can know what is changed.…

Codeforces Round #698 Nezzar and Symmetric Array - Construction

By xloypaypa in codeforces on 03 Apr 2021

First let's choose any two numbers, $a_i$ and $a_j$. So There must be $-a_i$ and $-a_j$ in $a$. And so there is no need to consider the symbol. Let's assume $a_i>0,\ a_j>0$.…

Codeforces Round #698 Nezzar and Lucky Number - Search by brute force

By xloypaypa in codeforces on 03 Apr 2021

Let current query is x, let remian=x mod d. If x>=10d}+remain. Then there must be a solution for x.…

Codeforces Round #702 Div3 Solution

By xloypaypa in codeforces on 01 Apr 2021

Solutions for all problems in Codeforces Round #702 Div3.…

Codeforces Round #708 Div2 Solution

By xloypaypa in codeforces on 01 Apr 2021

Solutions for C2, D and E2 in Codeforces Round #708 Div2.…

Codeforces Round #708 Square-free division - DP

By xloypaypa in codeforces on 01 Apr 2021

There is only one case to make product is a perfect square: For each number, the number of occurrences of prime factors is count[i]. if any different $i$, $j$ to make count[i]=count[j].…

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