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 thinking.. Ishu RajoraIET Lucknow
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...
The CodeChef problem setting portal is still not a very user friendly system. And to use it you have to somehow learn how to do it. There is a guide for it... accessible only by a link you have to somehow get. Also this guide is written several years ago. Yeah. Actually, there a...
Dominater069 Written editorials will be available for all ondiscuss.codechef.com. Pro users can find the editorials directly on the problem pages after the contest. The video editorials of the problems will be available only to Pro users. Also, if you have some original and engaging problem i...
CodeChef problem solution. Contribute to HunterMahmud/CodeChef development by creating an account on GitHub.
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
codechef problems 第二弹 一、Backup Functions 题面 One unavoidable problem with running a restaurant is that occasionally a menu item cannot be prepared. This can be caused by a variety of reasons such as missing ingredients or malfunctioning equipment. ...
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
原题来自:CODECHEF September Challenge 2015 REBXOR 给定一个含 N 个元素的数组 A,下标从 1 开始。请找出下面式子的最大值: (A[l1]⨁A[l1+1]⨁…⨁A[r1])+(A[l2]⨁A[l2+1]…⨁A[r2]),其中1≤l1≤r1<l2≤r2≤N,x⨁y 表示 x 和 y 的按位异或。