CodeChef transformed my thinking. It improved my problem-solving, introduced new algorithms, and taught me to optimize. The discussions and editorials broadened my approach, enhancing both coding and logical th
感觉codechef easy里面的都是智商题啊qwq。。 这题我们可以按位考虑。 如果当前为二进制下第$i$位,区间$(l, r)$中元素第$i$位含有$1$的有$x$个 很显然,当$x <= \frac{r - l + 1}{2}$时$0$的个数多,那么答案的第$i$位应为$1$,否则为$0$ 代码语言:javascript 代码运行次数:0 运行 AI代...
Joining us on the problem setting panel are: Setters: Aditya KumarAdityak2507Sahu, Ananyananymous507Dhamija, Anujanjkk01Gupta, MehulMehul_Gupta Dominater069 Written editorials will be available for all ondiscuss.codechef.com. Pro users can find the editorials directly on the problem pages after ...
Also, if you have some original and engaging problem ideas, and you’re interested in them being used in CodeChef's contests, you can share them here. Hope to see you participating. Good Luck! +56 yash_daga 22 months ago 15 Comments...
Note:If you are applying for becoming a new problem setter in CodeChef, and we don't know you personally and there isn't a problem setter of CodeChef who can vouch for your skills, then you must satisfy at least one of the below criteria for getting your problem setter application consi...
题面: 小明已经厌倦了做菜,做菜对他来说太简单了。这天,他决定要挑战一下自己。 他已经选择了 N 根意大利面,准备做一道菜。他还需要选出一根意大利面,使得这根意大利面可以和已有的 N 根面中的某两根构成一个三角形。选出的意大利面的长度必须在 [L, R] 的范围内。 请
题目链接:http://vjudge.net/problem/174774/origin 题意:给定一个长度为n的序列a[],序列的值不大于m,现在有k个询问,每个询问给定(l,r).让你求出max{|x − y| : Li ≤ x, y ≤ Ri and Ax = Ay}。即区间[L,R]中值相同时,位置差的最大值 ...
Code Issues Pull requests 📌 📚 Solution of competitive programming problems, code templates, Data Structures and Algorithms, hackathons, interviews and much more. javascript python google facebook algorithm algorithms datastructures leetcode cpp hackathon interview codechef competitive-programming hackerra...
This repository is dedicated to solving various coding problems and challenges across multiple platforms. Here, you will find solutions and code examples for problems from different sources(Codeforces, Codechef, AtCoder and LeetCode) - GitWizAmit/Problem
CodeChef - FNCS Chef and Churu(分块),https://vjudge.net/problem/CodeChef-FNCS题意思路用分块的方法,对每个函数进行分块,计算出该分块里每个数的个数,这样的话也就能很方便的计算出这个分块里所有数的和。用树状数组维护数组的话可以很方便的计算出某个区间内所有